If a : b : : c : d, prove that
(i) 2 a + 5 b 2 a − 5 b = 2 c + 5 d 2 c − 5 d . (ii) 5 a + 11 b 5 c + 11 d = 5 a − 11 b 5 c − 11 d . (iii) ( 2 a + 3 b ) ( 2 c − 3 d ) = ( 2 a − 3 b ) ( 2 c + 3 d ) . (iv) ( l a + m b ) : ( l c + m d ) : : ( l a − m b ) : ( l c − m d ) . \begin{array}{ll} \text{(i)} & \dfrac{2a + 5b}{2a - 5b} = \dfrac{2c + 5d}{2c - 5d}. \\[1em] \text{(ii)} & \dfrac{5a + 11b}{5c + 11d} = \dfrac{5a - 11b}{5c - 11d}. \\[1em] \text{(iii)} & (2a + 3b)(2c - 3d) = (2a - 3b)(2c + 3d). \\[0.7em] \text{(iv)} & (la + mb) : (lc + md) : : (la - mb) : (lc - md). \end{array} (i) (ii) (iii) (iv) 2 a − 5 b 2 a + 5 b = 2 c − 5 d 2 c + 5 d . 5 c + 11 d 5 a + 11 b = 5 c − 11 d 5 a − 11 b . ( 2 a + 3 b ) ( 2 c − 3 d ) = ( 2 a − 3 b ) ( 2 c + 3 d ) . ( l a + mb ) : ( l c + m d ) :: ( l a − mb ) : ( l c − m d ) .
Answer
(i) Given, a : b : : c : d,
⇒ a b = c d \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[0.5em] ⇒ b a = d c
Multiplying the equation by 2 5 \dfrac{2}{5} 5 2
⇒ 2 a 5 b = 2 c 5 d \Rightarrow \dfrac{2a}{5b} = \dfrac{2c}{5d} \\[0.5em] ⇒ 5 b 2 a = 5 d 2 c
By componendo and dividendo,
⇒ 2 a + 5 b 2 a − 5 b = 2 c + 5 d 2 c − 5 d \Rightarrow \dfrac{2a + 5b}{2a - 5b} = \dfrac{2c + 5d}{2c - 5d} \\[0.5em] ⇒ 2 a − 5 b 2 a + 5 b = 2 c − 5 d 2 c + 5 d
Hence, proved that 2 a + 5 b 2 a − 5 b = 2 c + 5 d 2 c − 5 d . \dfrac{2a + 5b}{2a - 5b} = \dfrac{2c + 5d}{2c - 5d}. 2 a − 5 b 2 a + 5 b = 2 c − 5 d 2 c + 5 d .
(ii) Given, a : b : : c : d,
⇒ a b = c d \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[0.5em] ⇒ b a = d c
On multiplying the equation by 5 11 \dfrac{5}{11} 11 5
⇒ 5 a 11 b = 5 c 11 d \Rightarrow \dfrac{5a}{11b} = \dfrac{5c}{11d} \\[0.5em] ⇒ 11 b 5 a = 11 d 5 c
By componendo and dividendo,
⇒ 5 a + 11 b 5 a − 11 b = 5 c + 11 d 5 c − 11 d \Rightarrow \dfrac{5a + 11b}{5a - 11b} = \dfrac{5c + 11d}{5c - 11d} \\[0.5em] ⇒ 5 a − 11 b 5 a + 11 b = 5 c − 11 d 5 c + 11 d
By alternendo,
⇒ 5 a + 11 b 5 c + 11 d = 5 a − 11 b 5 c − 11 d \Rightarrow \dfrac{5a + 11b}{5c + 11d} = \dfrac{5a - 11b}{5c - 11d} ⇒ 5 c + 11 d 5 a + 11 b = 5 c − 11 d 5 a − 11 b
Hence, proved that 5 a + 11 b 5 c + 11 d = 5 a − 11 b 5 c − 11 d . \dfrac{5a + 11b}{5c + 11d} = \dfrac{5a - 11b}{5c - 11d}. 5 c + 11 d 5 a + 11 b = 5 c − 11 d 5 a − 11 b .
(iii) Given, a : b : : c : d,
⇒ a b = c d \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[0.5em] ⇒ b a = d c
On multiplying equation by 2 3 \dfrac{2}{3} 3 2
⇒ 2 a 3 b = 2 c 3 d \Rightarrow \dfrac{2a}{3b} = \dfrac{2c}{3d} \\[0.5em] ⇒ 3 b 2 a = 3 d 2 c
By componendo and dividendo,
⇒ 2 a + 3 b 2 a − 3 b = 2 c + 3 d 2 c − 3 d \Rightarrow \dfrac{2a + 3b}{2a - 3b} = \dfrac{2c + 3d}{2c - 3d} \\[0.5em] ⇒ 2 a − 3 b 2 a + 3 b = 2 c − 3 d 2 c + 3 d
On cross multiplication,
⇒ ( 2 a + 3 b ) ( 2 c − 3 d ) = ( 2 c + 3 d ) ( 2 a − 3 b ) . \Rightarrow (2a + 3b)(2c - 3d) = (2c + 3d)(2a - 3b). ⇒ ( 2 a + 3 b ) ( 2 c − 3 d ) = ( 2 c + 3 d ) ( 2 a − 3 b ) .
Hence, proved that (2a + 3b)(2c - 3d) = (2a - 3b)(2c + 3d).
(iv) Given, a : b : : c : d,
⇒ a b = c d \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[0.5em] ⇒ b a = d c
On multiplying the equation by l m \dfrac{l}{m} m l
⇒ l a m b = l c m d \Rightarrow \dfrac{la}{mb} = \dfrac{lc}{md} \\[0.5em] ⇒ mb l a = m d l c
By componendo and dividendo,
⇒ l a + m b l a − m b = l c + m d l c − m d \Rightarrow \dfrac{la + mb}{la - mb} = \dfrac{lc + md}{lc - md} \\[0.5em] ⇒ l a − mb l a + mb = l c − m d l c + m d
By alternendo,
⇒ l a + m b l c + m d = l a − m b l c − m d ⇒ ( l a + m b ) : ( l c + m d ) : : ( l a − m b ) : ( l c − m d ) . \Rightarrow \dfrac{la + mb}{lc + md} = \dfrac{la - mb}{lc - md} \\[0.5em] \Rightarrow (la + mb) : (lc + md) : : (la - mb) : (lc - md). ⇒ l c + m d l a + mb = l c − m d l a − mb ⇒ ( l a + mb ) : ( l c + m d ) :: ( l a − mb ) : ( l c − m d ) .
Hence, proved that (la + mb) : (lc + md) : : (la - mb) : (lc - md).
If 5 x + 7 y 5 u + 7 v = 5 x − 7 y 5 u − 7 v , show that x y = u v . \dfrac{5x + 7y}{5u + 7v} = \dfrac{5x - 7y}{5u - 7v}, \text{ show that } \dfrac{x}{y} = \dfrac{u}{v}. 5 u + 7 v 5 x + 7 y = 5 u − 7 v 5 x − 7 y , show that y x = v u .
Answer
Given,
5 x + 7 y 5 u + 7 v = 5 x − 7 y 5 u − 7 v \dfrac{5x + 7y}{5u + 7v} = \dfrac{5x - 7y}{5u - 7v} 5 u + 7 v 5 x + 7 y = 5 u − 7 v 5 x − 7 y
By alternendo,
⇒ 5 x + 7 y 5 x − 7 y = 5 u + 7 v 5 u − 7 v \Rightarrow \dfrac{5x + 7y}{5x - 7y} = \dfrac{5u + 7v}{5u - 7v} \\[0.5em] ⇒ 5 x − 7 y 5 x + 7 y = 5 u − 7 v 5 u + 7 v
By componendo & dividendo,
⇒ 5 x + 7 y + 5 x − 7 y 5 x + 7 y − 5 x + 7 y = 5 u + 7 v + 5 u − 7 v 5 u + 7 v − 5 u + 7 v ⇒ 10 x 14 y = 10 u 14 v \Rightarrow \dfrac{5x + 7y + 5x - 7y}{5x + 7y - 5x + 7y} = \dfrac{5u + 7v + 5u - 7v}{5u + 7v - 5u + 7v} \\[0.5em] \Rightarrow \dfrac{10x}{14y} = \dfrac{10u}{14v} \\[0.5em] ⇒ 5 x + 7 y − 5 x + 7 y 5 x + 7 y + 5 x − 7 y = 5 u + 7 v − 5 u + 7 v 5 u + 7 v + 5 u − 7 v ⇒ 14 y 10 x = 14 v 10 u
On dividing equation by 10 14 \dfrac{10}{14} 14 10
⇒ x y = u v \Rightarrow \dfrac{x}{y} = \dfrac{u}{v} \\[0.5em] ⇒ y x = v u
Hence, proved that x y = u v . \dfrac{x}{y} = \dfrac{u}{v}. y x = v u .
8 a − 5 b 8 c − 5 d = 8 a + 5 b 8 c + 5 d , prove that a b = c d . \dfrac{8a - 5b}{8c - 5d} = \dfrac{8a + 5b}{8c + 5d}, \text{ prove that } \dfrac{a}{b} = \dfrac{c}{d}. 8 c − 5 d 8 a − 5 b = 8 c + 5 d 8 a + 5 b , prove that b a = d c .
Answer
Given,
8 a − 5 b 8 c − 5 d = 8 a + 5 b 8 c + 5 d \dfrac{8a - 5b}{8c - 5d} = \dfrac{8a + 5b}{8c + 5d} \\[0.5em] 8 c − 5 d 8 a − 5 b = 8 c + 5 d 8 a + 5 b
By alternendo,
⇒ 8 a − 5 b 8 a + 5 b = 8 c − 5 d 8 c + 5 d \Rightarrow \dfrac{8a - 5b}{8a + 5b} = \dfrac{8c - 5d}{8c + 5d} \\[0.5em] ⇒ 8 a + 5 b 8 a − 5 b = 8 c + 5 d 8 c − 5 d
By componendo & dividendo,
⇒ 8 a − 5 b + 8 a + 5 b 8 a − 5 b − 8 a − 5 b = 8 c − 5 d + 8 c + 5 d 8 c − 5 d − 8 c − 5 d ⇒ − 16 a 10 b = − 16 c 10 d \Rightarrow \dfrac{8a - 5b + 8a + 5b}{8a - 5b - 8a - 5b} = \dfrac{8c - 5d + 8c + 5d}{8c - 5d - 8c - 5d} \\[1em] \Rightarrow -\dfrac{16a}{10b} = -\dfrac{16c}{10d} ⇒ 8 a − 5 b − 8 a − 5 b 8 a − 5 b + 8 a + 5 b = 8 c − 5 d − 8 c − 5 d 8 c − 5 d + 8 c + 5 d ⇒ − 10 b 16 a = − 10 d 16 c
On dividing the equation by − 16 10 -\dfrac{16}{10} − 10 16
⇒ a b = c d \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[0.5em] ⇒ b a = d c
Hence, proved that a b = c d . \dfrac{a}{b} = \dfrac{c}{d}. b a = d c .
If (4a + 5b)(4c - 5d) = (4a - 5b)(4c + 5d), prove that a, b, c, d are in proportion.
Answer
Given, (4a + 5b)(4c - 5d) = (4a - 5b)(4c + 5d).
On cross-multiplication,
⇒ 4 a + 5 b 4 a − 5 b = 4 c + 5 d 4 c − 5 d \Rightarrow \dfrac{4a + 5b}{4a - 5b} = \dfrac{4c + 5d}{4c - 5d} \\[0.5em] ⇒ 4 a − 5 b 4 a + 5 b = 4 c − 5 d 4 c + 5 d
By componendo and dividendo,
⇒ 4 a + 5 b + 4 a − 5 b 4 a + 5 b − 4 a + 5 b = 4 c + 5 d + 4 c − 5 d 4 c + 5 d − 4 c + 5 d ⇒ 8 a 10 b = 8 c 10 d \Rightarrow \dfrac{4a + 5b + 4a - 5b}{4a + 5b - 4a + 5b} = \dfrac{4c + 5d + 4c - 5d}{4c + 5d - 4c + 5d} \\[0.5em] \Rightarrow \dfrac{8a}{10b} = \dfrac{8c}{10d} \\[0.5em] ⇒ 4 a + 5 b − 4 a + 5 b 4 a + 5 b + 4 a − 5 b = 4 c + 5 d − 4 c + 5 d 4 c + 5 d + 4 c − 5 d ⇒ 10 b 8 a = 10 d 8 c
On dividing the equation by 8 10 \dfrac{8}{10} 10 8
⇒ a b = c d ⇒ a : b : : c : d . \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[0.5em] \Rightarrow a : b : : c : d. ⇒ b a = d c ⇒ a : b :: c : d .
Hence, proved that a, b, c, d are in proportion.
If (pa + qb) : (pc + qd) : : (pa - qb) : (pc - qd), prove that a : b : : c : d.
Answer
Given, (pa + qb) : (pc + qd) : : (pa - qb) : (pc - qd).
⇒ p a + q b p c + q d = p a − q b p c − q d \Rightarrow \dfrac{pa + qb}{pc + qd} = \dfrac{pa - qb}{pc - qd} \\[0.5em] ⇒ p c + q d p a + q b = p c − q d p a − q b
By alternendo,
⇒ p a + q b p a − q b = p c + q d p c − q d \Rightarrow \dfrac{pa + qb}{pa - qb} = \dfrac{pc + qd}{pc - qd} \\[0.5em] ⇒ p a − q b p a + q b = p c − q d p c + q d
By componendo and dividendo,
⇒ p a + q b + p a − q b p a + q b − p a + q b = p c + q d + p c − q d p c + q d − p c + q d ⇒ 2 p a 2 q b = 2 p c 2 q d \Rightarrow \dfrac{pa + qb + pa - qb}{pa + qb - pa + qb} = \dfrac{pc + qd + pc - qd}{pc + qd - pc + qd} \\[0.5em] \Rightarrow \dfrac{2pa}{2qb} = \dfrac{2pc}{2qd} \\[0.5em] ⇒ p a + q b − p a + q b p a + q b + p a − q b = p c + q d − p c + q d p c + q d + p c − q d ⇒ 2 q b 2 p a = 2 q d 2 p c
On dividing the equation by 2 p 2 q \dfrac{2p}{2q} 2 q 2 p
⇒ a b = c d ⇒ a : b : : c : d . \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[0.5em] \Rightarrow a : b : : c : d. ⇒ b a = d c ⇒ a : b :: c : d .
Hence, proved that a, b, c, d are in proportion.
If (ma + nb) : b : : (mc + nd) : d, prove that a, b, c, d are in proportion.
Answer
Given, (ma + nb) : b : : (mc + nd) : d.
⇒ ( m a + n b ) b = ( m c + n d ) d ⇒ d ( m a + n b ) = b ( m c + n d ) ⇒ m a d + n b d = b m c + b n d ⇒ m a d = b m c \Rightarrow \dfrac{(ma + nb)}{b} = \dfrac{(mc + nd)}{d} \\[0.5em] \Rightarrow d(ma + nb) = b(mc + nd) \\[0.5em] \Rightarrow mad + nbd = bmc + bnd \\[0.5em] \Rightarrow mad = bmc \\[0.5em] ⇒ b ( ma + nb ) = d ( m c + n d ) ⇒ d ( ma + nb ) = b ( m c + n d ) ⇒ ma d + nb d = bm c + bn d ⇒ ma d = bm c
On dividing equation by m,
⇒ a d = b c ⇒ a b = c d ⇒ a : b : : c : d . \Rightarrow ad = bc \\[0.5em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[0.5em] \Rightarrow a : b : : c : d. ⇒ a d = b c ⇒ b a = d c ⇒ a : b :: c : d .
Hence, proved that a, b, c, d are in proportion.
If (11a2 + 13b2 )(11c2 - 13d2 ) = (11a2 - 13b2 )(11c2 + 13d2 ), prove that a : b : : c : d.
Answer
Given, (11a2 + 13b2 )(11c2 - 13d2 ) = (11a2 - 13b2 )(11c2 + 13d2 ).
On cross-multiplication,
⇒ 11 a 2 + 13 b 2 11 a 2 − 13 b 2 = 11 c 2 + 13 d 2 11 c 2 − 13 d 2 \Rightarrow \dfrac{11a^2 + 13b^2}{11a^2 - 13b^2} = \dfrac{11c^2 + 13d^2}{11c^2 - 13d^2} \\[0.5em] ⇒ 11 a 2 − 13 b 2 11 a 2 + 13 b 2 = 11 c 2 − 13 d 2 11 c 2 + 13 d 2
By componendo and dividendo,
⇒ 11 a 2 + 13 b 2 + 11 a 2 − 13 b 2 11 a 2 + 13 b 2 − 11 a 2 + 13 b 2 = 11 c 2 + 13 d 2 + 11 c 2 − 13 d 2 11 c 2 + 13 d 2 − 11 c 2 + 13 d 2 ⇒ 22 a 2 26 b 2 = 22 c 2 26 d 2 \Rightarrow \dfrac{11a^2 + 13b^2 + 11a^2 - 13b^2}{11a^2 + 13b^2 - 11a^2 + 13b^2} = \dfrac{11c^2 + 13d^2 + 11c^2 - 13d^2}{11c^2 + 13d^2 - 11c^2 + 13d^2} \\[0.5em] \Rightarrow \dfrac{22a^2}{26b^2} = \dfrac{22c^2}{26d^2} \\[0.5em] ⇒ 11 a 2 + 13 b 2 − 11 a 2 + 13 b 2 11 a 2 + 13 b 2 + 11 a 2 − 13 b 2 = 11 c 2 + 13 d 2 − 11 c 2 + 13 d 2 11 c 2 + 13 d 2 + 11 c 2 − 13 d 2 ⇒ 26 b 2 22 a 2 = 26 d 2 22 c 2
On dividing the equation by 22 26 \dfrac{22}{26} 26 22
⇒ a 2 b 2 = c 2 d 2 ⇒ ( a 2 b 2 ) = ( c 2 d 2 ) ⇒ a b = c d ⇒ a : b : : c : d . \Rightarrow \dfrac{a^2}{b^2} = \dfrac{c^2}{d^2} \\[0.5em] \Rightarrow \sqrt{\big(\dfrac{a^2}{b^2}\big)} = \sqrt{\big(\dfrac{c^2}{d^2}\big)} \\[0.5em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d} \\[0.5em] \Rightarrow a : b : : c : d. ⇒ b 2 a 2 = d 2 c 2 ⇒ ( b 2 a 2 ) = ( d 2 c 2 ) ⇒ b a = d c ⇒ a : b :: c : d .
Hence, proved that a, b, c, d are in proportion.
If x = 2 a b a + b \dfrac{2ab}{a + b} a + b 2 ab x + a x − a + x + b x − b \dfrac{x + a}{x - a} + \dfrac{x + b}{x - b} x − a x + a + x − b x + b
Answer
Given,
x = 2 a b a + b x a = 2 a b a + b a ∴ x a = 2 b a + b x = \dfrac{2ab}{a + b} \\[1em] \dfrac{x}{a} = \dfrac{\dfrac{2ab}{a + b}}{a} \\[1em] \therefore \dfrac{x}{a} = \dfrac{2b}{a + b} x = a + b 2 ab a x = a a + b 2 ab ∴ a x = a + b 2 b
By componendo and dividendo,
⇒ x + a x − a = 2 b + a + b 2 b − a − b ⇒ x + a x − a = 3 b + a b − a [....Eq 1] x b = 2 a b a + b b ∴ x b = 2 a a + b \Rightarrow \dfrac{x + a}{x - a} = \dfrac{2b + a + b}{2b - a - b } \\[0.5em] \Rightarrow \dfrac{x + a}{x - a} = \dfrac{3b + a}{b - a} \qquad \text{[....Eq 1]} \\[1.5em] \dfrac{x}{b} = \dfrac{\dfrac{2ab}{a + b}}{b} \\[1em] \therefore \dfrac{x}{b} = \dfrac{2a}{a + b} \\[1em] ⇒ x − a x + a = 2 b − a − b 2 b + a + b ⇒ x − a x + a = b − a 3 b + a [....Eq 1] b x = b a + b 2 ab ∴ b x = a + b 2 a
By componendo and dividendo,
⇒ x + b x − b = 2 a + a + b 2 a − a − b ⇒ x + b x − b = 3 a + b a − b [....Eq 2] \Rightarrow \dfrac{x + b}{x - b} = \dfrac{2a + a + b}{2a - a - b} \\[0.5em] \Rightarrow \dfrac{x + b}{x - b} = \dfrac{3a + b}{a - b} \qquad \text{[....Eq 2]} ⇒ x − b x + b = 2 a − a − b 2 a + a + b ⇒ x − b x + b = a − b 3 a + b [....Eq 2]
Adding Eq 1 and 2,
⇒ x + a x − a + x + b x − b = 3 b + a b − a + 3 a + b a − b ⇒ x + a x − a + x + b x − b = 3 b + a b − a − 3 a + b b − a ⇒ x + a x − a + x + b x − b = 3 b − b + a − 3 a b − a ⇒ x + a x − a + x + b x − b = 2 b − 2 a b − a ⇒ x + a x − a + x + b x − b = 2 ( b − a ) b − a = 2 \Rightarrow \dfrac{x + a}{x - a} + \dfrac{x + b}{x - b} = \dfrac{3b + a}{b - a} + \dfrac{3a + b}{a - b} \\[1em] \Rightarrow \dfrac{x + a}{x - a} + \dfrac{x + b}{x - b} = \dfrac{3b + a}{b - a} - \dfrac{3a + b}{b - a} \\[1em] \Rightarrow \dfrac{x + a}{x - a} + \dfrac{x + b}{x - b} = \dfrac{3b - b + a - 3a}{b - a} \\[1em] \Rightarrow \dfrac{x + a}{x - a} + \dfrac{x + b}{x - b} = \dfrac{2b - 2a}{b - a} \\[1em] \Rightarrow \dfrac{x + a}{x - a} + \dfrac{x + b}{x - b} = \dfrac{2(b - a)}{b - a} = 2 \\[1em] ⇒ x − a x + a + x − b x + b = b − a 3 b + a + a − b 3 a + b ⇒ x − a x + a + x − b x + b = b − a 3 b + a − b − a 3 a + b ⇒ x − a x + a + x − b x + b = b − a 3 b − b + a − 3 a ⇒ x − a x + a + x − b x + b = b − a 2 b − 2 a ⇒ x − a x + a + x − b x + b = b − a 2 ( b − a ) = 2
Hence, the required value is 2.
If x = 8 a b a + b , \dfrac{8ab}{a + b}, a + b 8 ab ,
x + 4 a x − 4 a + x + 4 b x − 4 b . \dfrac{x + 4a}{x - 4a} + \dfrac{x + 4b}{x - 4b}. x − 4 a x + 4 a + x − 4 b x + 4 b .
Answer
Given,
x = 8 a b a + b x 4 a = 8 a b a + b 4 a ∴ x 4 a = 2 b a + b x = \dfrac{8ab}{a + b} \\[1em] \dfrac{x}{4a} = \dfrac{\dfrac{8ab}{a + b}}{4a} \\[1em] \therefore \dfrac{x}{4a} = \dfrac{2b}{a + b} x = a + b 8 ab 4 a x = 4 a a + b 8 ab ∴ 4 a x = a + b 2 b
By componendo and dividendo, ⇒ x + 4 a x − 4 a = 2 b + a + b 2 b − a − b ⇒ x + 4 a x − 4 a = 3 b + a b − a [....Eq 1] x 4 b = 8 a b a + b 4 b ∴ x 4 b = 2 a a + b \Rightarrow \dfrac{x + 4a}{x - 4a} = \dfrac{2b + a + b}{2b - a - b } \\[0.5em] \Rightarrow \dfrac{x + 4a}{x - 4a} = \dfrac{3b + a}{b - a} \qquad \text{[....Eq 1]} \\[1.5em] \dfrac{x}{4b} = \dfrac{\dfrac{8ab}{a + b}}{4b} \\[1em] \therefore \dfrac{x}{4b} = \dfrac{2a}{a + b} ⇒ x − 4 a x + 4 a = 2 b − a − b 2 b + a + b ⇒ x − 4 a x + 4 a = b − a 3 b + a [....Eq 1] 4 b x = 4 b a + b 8 ab ∴ 4 b x = a + b 2 a
By componendo and dividendo,
⇒ x + 4 b x − 4 b = 2 a + a + b 2 a − a − b ⇒ x + 4 b x − 4 b = 3 a + b a − b [....Eq 2] \Rightarrow \dfrac{x + 4b}{x - 4b} = \dfrac{2a + a + b}{2a - a - b} \\[0.5em] \Rightarrow \dfrac{x + 4b}{x - 4b} = \dfrac{3a + b}{a - b} \qquad \text{[....Eq 2]} ⇒ x − 4 b x + 4 b = 2 a − a − b 2 a + a + b ⇒ x − 4 b x + 4 b = a − b 3 a + b [....Eq 2]
Adding Eq 1 and 2,
⇒ x + 4 a x − 4 a + x + 4 b x − 4 b = 3 b + a b − a + 3 a + b a − b ⇒ x + 4 a x − 4 a + x + 4 b x − 4 b = 3 b + a b − a − 3 a + b b − a ⇒ x + 4 a x − 4 a + x + 4 b x − 4 b = 3 b − b + a − 3 a b − a ⇒ x + 4 a x − 4 a + x + 4 b x − 4 b = 2 b − 2 a b − a ⇒ x + 4 a x − 4 a + x + 4 b x − 4 b = 2 ( b − a ) b − a = 2 \Rightarrow \dfrac{x + 4a}{x - 4a} + \dfrac{x + 4b}{x - 4b} = \dfrac{3b + a}{b - a} + \dfrac{3a + b}{a - b} \\[1em] \Rightarrow \dfrac{x + 4a}{x - 4a} + \dfrac{x + 4b}{x - 4b} = \dfrac{3b + a}{b - a} - \dfrac{3a + b}{b - a} \\[1em] \Rightarrow \dfrac{x + 4a}{x - 4a} + \dfrac{x + 4b}{x - 4b} = \dfrac{3b - b + a - 3a}{b - a} \\[1em] \Rightarrow \dfrac{x + 4a}{x - 4a} + \dfrac{x + 4b}{x - 4b} = \dfrac{2b - 2a}{b - a} \\[1em] \Rightarrow \dfrac{x + 4a}{x - 4a} + \dfrac{x + 4b}{x - 4b} = \dfrac{2(b - a)}{b - a} = 2 \\[1em] ⇒ x − 4 a x + 4 a + x − 4 b x + 4 b = b − a 3 b + a + a − b 3 a + b ⇒ x − 4 a x + 4 a + x − 4 b x + 4 b = b − a 3 b + a − b − a 3 a + b ⇒ x − 4 a x + 4 a + x − 4 b x + 4 b = b − a 3 b − b + a − 3 a ⇒ x − 4 a x + 4 a + x − 4 b x + 4 b = b − a 2 b − 2 a ⇒ x − 4 a x + 4 a + x − 4 b x + 4 b = b − a 2 ( b − a ) = 2
Hence, the required value is 2.
If x = 4 6 2 + 3 \dfrac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}} 2 + 3 4 6
x + 2 2 x − 2 2 + x + 2 3 x − 2 3 \dfrac{x + 2\sqrt{2}}{x - 2\sqrt{2}} + \dfrac{x + 2\sqrt{3}}{x - 2\sqrt{3}} x − 2 2 x + 2 2 + x − 2 3 x + 2 3
Answer
Given,
x = 4 6 2 + 3 x 2 2 = 4 6 2 + 3 2 2 ∴ x 2 2 = 2 3 2 + 3 x = \dfrac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}} \\[1em] \dfrac{x}{2\sqrt{2}} = \dfrac{\dfrac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}}}{2\sqrt{2}} \\[1em] \therefore \dfrac{x}{2\sqrt{2}} = \dfrac{2\sqrt{3}}{\sqrt{2} + \sqrt{3}} x = 2 + 3 4 6 2 2 x = 2 2 2 + 3 4 6 ∴ 2 2 x = 2 + 3 2 3
By componendo and dividendo,
⇒ x + 2 2 x − 2 2 = 2 3 + 2 + 3 2 3 − 2 − 3 ⇒ x + 2 2 x − 2 2 = 3 3 + 2 3 − 2 [....Eq 1] x 2 3 = 4 6 2 + 3 2 3 ∴ x 2 3 = 2 2 2 + 3 \Rightarrow \dfrac{x + 2\sqrt{2}}{x - 2\sqrt{2}} = \dfrac{2\sqrt{3} + \sqrt{2} + \sqrt{3}}{2\sqrt{3} - \sqrt{2} - \sqrt{3}} \\[1em] \Rightarrow \dfrac{x + 2\sqrt{2}}{x - 2\sqrt{2}} = \dfrac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \qquad \text{[....Eq 1]} \\[1.5em] \dfrac{x}{2\sqrt{3}} = \dfrac{\dfrac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}}}{2\sqrt{3}} \\[1em] \therefore \dfrac{x}{2\sqrt{3}} = \dfrac{2\sqrt{2}}{\sqrt{2} + \sqrt{3}} ⇒ x − 2 2 x + 2 2 = 2 3 − 2 − 3 2 3 + 2 + 3 ⇒ x − 2 2 x + 2 2 = 3 − 2 3 3 + 2 [....Eq 1] 2 3 x = 2 3 2 + 3 4 6 ∴ 2 3 x = 2 + 3 2 2
By componendo and dividendo,
⇒ x + 2 3 x − 2 3 = 2 2 + 2 + 3 2 2 − 2 − 3 ⇒ x + 2 3 x − 2 3 = 3 2 + 3 2 − 3 [....Eq 2] \Rightarrow \dfrac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \dfrac{2\sqrt{2} + \sqrt{2} + \sqrt{3}}{2\sqrt{2} - \sqrt{2} - \sqrt{3}} \\[1em] \Rightarrow \dfrac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \dfrac{3\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} \qquad \text{[....Eq 2]} ⇒ x − 2 3 x + 2 3 = 2 2 − 2 − 3 2 2 + 2 + 3 ⇒ x − 2 3 x + 2 3 = 2 − 3 3 2 + 3 [....Eq 2]
Adding Eq 1 and 2,
⇒ x + 2 2 x − 2 2 + x + 2 3 x − 2 3 = 3 3 + 2 3 − 2 + 3 2 + 3 2 − 3 ⇒ x + 2 2 x − 2 2 + x + 2 3 x − 2 3 = 3 3 + 2 3 − 2 − 3 2 + 3 3 − 2 ⇒ x + 2 2 x − 2 2 + x + 2 3 x − 2 3 = 3 3 + 2 − 3 2 − 3 3 − 2 ⇒ x + 2 2 x − 2 2 + x + 2 3 x − 2 3 = 2 3 − 2 2 3 − 2 ⇒ x + 2 2 x − 2 2 + x + 2 3 x − 2 3 = 2 ( 3 − 2 ) 3 − 2 = 2. \Rightarrow \dfrac{x + 2\sqrt{2}}{x - 2\sqrt{2}} + \dfrac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \dfrac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} + \dfrac{3\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} \\[1em] \Rightarrow \dfrac{x + 2\sqrt{2}}{x - 2\sqrt{2}} + \dfrac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \dfrac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} - \dfrac{3\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2}} \\[1em] \Rightarrow \dfrac{x + 2\sqrt{2}}{x - 2\sqrt{2}} + \dfrac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \dfrac{3\sqrt{3} + \sqrt{2} - 3\sqrt{2} - \sqrt{3}}{\sqrt{3} - \sqrt{2}} \\[1em] \Rightarrow \dfrac{x + 2\sqrt{2}}{x - 2\sqrt{2}} + \dfrac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \dfrac{2\sqrt{3} - 2\sqrt{2}}{\sqrt{3} - \sqrt{2}} \\[1em] \Rightarrow \dfrac{x + 2\sqrt{2}}{x - 2\sqrt{2}} + \dfrac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \dfrac{2(\sqrt{3} - \sqrt{2})}{\sqrt{3} - \sqrt{2}} = 2. \\[1em] ⇒ x − 2 2 x + 2 2 + x − 2 3 x + 2 3 = 3 − 2 3 3 + 2 + 2 − 3 3 2 + 3 ⇒ x − 2 2 x + 2 2 + x − 2 3 x + 2 3 = 3 − 2 3 3 + 2 − 3 − 2 3 2 + 3 ⇒ x − 2 2 x + 2 2 + x − 2 3 x + 2 3 = 3 − 2 3 3 + 2 − 3 2 − 3 ⇒ x − 2 2 x + 2 2 + x − 2 3 x + 2 3 = 3 − 2 2 3 − 2 2 ⇒ x − 2 2 x + 2 2 + x − 2 3 x + 2 3 = 3 − 2 2 ( 3 − 2 ) = 2.
Hence, the required value is 2.
Using properties of proportion, find x from the following equations :
(i) 2 − x + 2 + x 2 − x − 2 + x = 3 (ii) x + 4 + x − 10 x + 4 + x − 10 = 5 2 (iii) 1 + x + 1 − x 1 + x − 1 − x = a b (iv) 5 x + 2 x − 6 5 x − 2 x − 6 = 4 (v) a + x + a − x a + x − a − x = c d (vi) a + a 2 − 2 a x a − a 2 − 2 a x = b \begin{matrix} \text{(i)} & \dfrac{\sqrt{2 - x} + \sqrt{2 + x}}{\sqrt{2 - x} - \sqrt{2 + x}} = 3 \\[2em] \text{(ii)} & \dfrac{\sqrt{x + 4} + \sqrt{x - 10}}{\sqrt{x + 4} + \sqrt{x - 10}} = \dfrac{5}{2} \\[2em] \text{(iii)} & \dfrac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} - \sqrt{1 - x}} = \dfrac{a}{b} \\[2em] \text{(iv)} & \dfrac{\sqrt{5x} + \sqrt{2x - 6}}{\sqrt{5x} - \sqrt{2x - 6}} = 4 \\[2em] \text{(v)} & \dfrac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = \dfrac{c}{d} \\[2em] \text{(vi)} & \dfrac{a + \sqrt{a^2 - 2ax}}{a - \sqrt{a^2 - 2ax}} = b \end{matrix} (i) (ii) (iii) (iv) (v) (vi) 2 − x − 2 + x 2 − x + 2 + x = 3 x + 4 + x − 10 x + 4 + x − 10 = 2 5 1 + x − 1 − x 1 + x + 1 − x = b a 5 x − 2 x − 6 5 x + 2 x − 6 = 4 a + x − a − x a + x + a − x = d c a − a 2 − 2 a x a + a 2 − 2 a x = b
Answer
(i) Given,
2 − x + 2 + x 2 − x − 2 + x = 3 1 \dfrac{\sqrt{2 - x} + \sqrt{2 + x}}{\sqrt{2 - x} - \sqrt{2 + x}} = \dfrac{3}{1} 2 − x − 2 + x 2 − x + 2 + x = 1 3
Applying componendo and dividendo,
⇒ 2 − x + 2 + x + 2 − x − 2 + x 2 − x + 2 + x − 2 − x + 2 + x = 3 + 1 3 − 1 ⇒ 2 2 − x 2 2 + x = 4 2 ⇒ 2 − x 2 + x = 2 1 \Rightarrow\dfrac{\sqrt{2 - x} + \sqrt{2 + x} + \sqrt{2 - x} - \sqrt{2 + x}}{\sqrt{2 - x} + \sqrt{2 + x} - \sqrt{2 - x} + \sqrt{2 + x}} = \dfrac{3 + 1}{3 - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{2 - x}}{2\sqrt{2 + x}} = \dfrac{4}{2} \\[1em] \Rightarrow \dfrac{\sqrt{2 - x}}{\sqrt{2 + x}} = \dfrac{2}{1} \\[1em] ⇒ 2 − x + 2 + x − 2 − x + 2 + x 2 − x + 2 + x + 2 − x − 2 + x = 3 − 1 3 + 1 ⇒ 2 2 + x 2 2 − x = 2 4 ⇒ 2 + x 2 − x = 1 2
Squaring both sides we get,
⇒ 2 − x 2 + x = 4 1 ⇒ ( 2 − x ) = 4 ( 2 + x ) ⇒ 2 − x = 8 + 4 x ⇒ 5 x = − 6 ⇒ x = − 6 5 . \Rightarrow \dfrac{2 - x}{2 + x} = \dfrac{4}{1} \\[0.5em] \Rightarrow (2 - x) = 4(2 + x) \\[0.5em] \Rightarrow 2 - x = 8 + 4x \\[0.5em] \Rightarrow 5x = -6 \\[0.5em] \Rightarrow x = -\dfrac{6}{5}. ⇒ 2 + x 2 − x = 1 4 ⇒ ( 2 − x ) = 4 ( 2 + x ) ⇒ 2 − x = 8 + 4 x ⇒ 5 x = − 6 ⇒ x = − 5 6 .
Hence, the value of x = − 6 5 . -\dfrac{6}{5}. − 5 6 .
(ii) Given,
x + 4 + x − 10 x + 4 − x − 10 = 5 2 \dfrac{\sqrt{x + 4} + \sqrt{x - 10}}{\sqrt{x + 4} - \sqrt{x - 10}} = \dfrac{5}{2} x + 4 − x − 10 x + 4 + x − 10 = 2 5
Applying componendo and dividendo,
⇒ x + 4 + x − 10 + x + 4 − x − 10 x + 4 + x − 10 − x − 4 + x − 10 = 5 + 2 5 − 2 ⇒ 2 x + 4 2 x − 10 = 7 3 ⇒ x + 4 x − 10 = 7 3 \Rightarrow\dfrac{\sqrt{x + 4} + \sqrt{x - 10} + \sqrt{x + 4} - \sqrt{x - 10}}{\sqrt{x + 4} + \sqrt{x - 10} - \sqrt{x - 4} + \sqrt{x - 10}} = \dfrac{5 + 2}{5 - 2} \\[1em] \Rightarrow \dfrac{2\sqrt{x + 4}}{2\sqrt{x - 10}} = \dfrac{7}{3} \\[1em] \Rightarrow \dfrac{\sqrt{x + 4}}{\sqrt{x - 10}} = \dfrac{7}{3} ⇒ x + 4 + x − 10 − x − 4 + x − 10 x + 4 + x − 10 + x + 4 − x − 10 = 5 − 2 5 + 2 ⇒ 2 x − 10 2 x + 4 = 3 7 ⇒ x − 10 x + 4 = 3 7
Squaring both sides,
⇒ x + 4 x − 10 = 49 9 ⇒ 9 ( x + 4 ) = 49 ( x − 10 ) ⇒ 9 x + 36 = 49 x − 490 ⇒ 40 x = 526 ⇒ x = 526 40 ⇒ x = 263 20 . \Rightarrow \dfrac{x + 4}{x - 10} = \dfrac{49}{9} \\[1em] \Rightarrow 9(x + 4) = 49(x - 10) \\[1em] \Rightarrow 9x + 36 = 49x - 490 \\[1em] \Rightarrow 40x = 526 \\[1em] \Rightarrow x = \dfrac{526}{40} \\[1em] \Rightarrow x = \dfrac{263}{20}. ⇒ x − 10 x + 4 = 9 49 ⇒ 9 ( x + 4 ) = 49 ( x − 10 ) ⇒ 9 x + 36 = 49 x − 490 ⇒ 40 x = 526 ⇒ x = 40 526 ⇒ x = 20 263 .
Hence, the value of x = 263 20 . \dfrac{263}{20}. 20 263 .
(iii) Given,
1 + x + 1 − x 1 + x − 1 − x = a b \dfrac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} - \sqrt{1 - x}} = \dfrac{a}{b} 1 + x − 1 − x 1 + x + 1 − x = b a
Applying componendo and dividendo,
⇒ 1 + x + 1 − x + 1 + x − 1 − x 1 + x + 1 − x − 1 + x + 1 − x = a + b a − b ⇒ 2 1 + x 2 1 − x = a + b a − b ⇒ 1 + x 1 − x = a + b a − b \Rightarrow\dfrac{\sqrt{1 + x} + \sqrt{1 - x} + \sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x} - \sqrt{1 + x} + \sqrt{1 - x}} = \dfrac{a + b}{a - b} \\[1em] \Rightarrow \dfrac{2\sqrt{1 + x}}{2\sqrt{1 - x}} = \dfrac{a + b}{a - b} \\[1em] \Rightarrow \dfrac{\sqrt{1 + x}}{{\sqrt{1 - x}}} = \dfrac{a + b}{a - b} ⇒ 1 + x + 1 − x − 1 + x + 1 − x 1 + x + 1 − x + 1 + x − 1 − x = a − b a + b ⇒ 2 1 − x 2 1 + x = a − b a + b ⇒ 1 − x 1 + x = a − b a + b
Squaring both sides we get,
⇒ 1 + x 1 − x = ( a + b ) 2 ( a − b ) 2 \Rightarrow \dfrac{1 + x}{1 - x} = \dfrac{(a + b)^2}{(a - b)^2} \\[0.5em] ⇒ 1 − x 1 + x = ( a − b ) 2 ( a + b ) 2
By componendo and dividendo,
⇒ 1 + x + 1 − x 1 + x − 1 + x = ( a + b ) 2 + ( a − b ) 2 ( a + b ) 2 − ( a − b ) 2 ⇒ 2 2 x = a 2 + 2 a b + b 2 + a 2 − 2 a b + b 2 a 2 + 2 a b + b 2 − a 2 + 2 a b − b 2 ⇒ 2 2 x = 2 a 2 + 2 b 2 2 a b + 2 a b ⇒ 2 2 x = 2 ( a 2 + b 2 ) 4 a b ⇒ 1 x = a 2 + b 2 2 a b ⇒ x = 2 a b a 2 + b 2 . \Rightarrow \dfrac{1 + x + 1 - x}{1 + x - 1 + x} = \dfrac{(a + b)^2 + (a - b)^2}{(a + b)^2 - (a - b)^2} \\[1em] \Rightarrow \dfrac{2}{2x} = \dfrac{a^2 + 2ab + b^2 + a^2 - 2ab + b^2}{a^2 + 2ab + b^2 - a^2 + 2ab - b^2} \\[1em] \Rightarrow \dfrac{2}{2x} = \dfrac{2a^2 + 2b^2 }{2ab + 2ab} \\[1em] \Rightarrow \dfrac{2}{2x} = \dfrac{2(a^2 + b^2)}{4ab} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{a^2 + b^2}{2ab} \\[1em] \Rightarrow x = \dfrac{2ab}{a^2 + b^2}. ⇒ 1 + x − 1 + x 1 + x + 1 − x = ( a + b ) 2 − ( a − b ) 2 ( a + b ) 2 + ( a − b ) 2 ⇒ 2 x 2 = a 2 + 2 ab + b 2 − a 2 + 2 ab − b 2 a 2 + 2 ab + b 2 + a 2 − 2 ab + b 2 ⇒ 2 x 2 = 2 ab + 2 ab 2 a 2 + 2 b 2 ⇒ 2 x 2 = 4 ab 2 ( a 2 + b 2 ) ⇒ x 1 = 2 ab a 2 + b 2 ⇒ x = a 2 + b 2 2 ab .
Hence, the value of x = 2 a b a 2 + b 2 . \dfrac{2ab}{a^2 + b^2}. a 2 + b 2 2 ab .
(iv) Given,
5 x + 2 x − 6 5 x − 2 x − 6 = 4. \dfrac{\sqrt{5x} + \sqrt{2x - 6}}{\sqrt{5x} - \sqrt{2x - 6}} = 4. 5 x − 2 x − 6 5 x + 2 x − 6 = 4.
By componendo and dividendo,
⇒ 5 x + 2 x − 6 + 5 x − 2 x − 6 5 x + 2 x − 6 − 5 x + 2 x − 6 = 4 + 1 4 − 1 ⇒ 2 5 x 2 2 x − 6 = 5 3 ⇒ 5 x 2 x − 6 = 5 3 \Rightarrow \dfrac{\sqrt{5x} + \sqrt{2x - 6} + \sqrt{5x} - \sqrt{2x - 6}}{\sqrt{5x} + \sqrt{2x - 6} - \sqrt{5x} + \sqrt{2x - 6}} = \dfrac{4 + 1}{4 - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{5x}}{2\sqrt{2x - 6}} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{\sqrt{5x}}{\sqrt{2x - 6}} = \dfrac{5}{3} ⇒ 5 x + 2 x − 6 − 5 x + 2 x − 6 5 x + 2 x − 6 + 5 x − 2 x − 6 = 4 − 1 4 + 1 ⇒ 2 2 x − 6 2 5 x = 3 5 ⇒ 2 x − 6 5 x = 3 5
Squaring both sides,
⇒ ( 5 x 2 x − 6 ) 2 = ( 5 3 ) 2 ⇒ 5 x 2 x − 6 = 25 9 ⇒ 5 x × 9 = 25 ( 2 x − 6 ) ⇒ 45 x = 50 x − 150 ⇒ 5 x = 150 ⇒ x = 30. \Rightarrow \Big(\dfrac{\sqrt{5x}}{\sqrt{2x - 6}}\Big)^2 = \Big(\dfrac{5}{3}\Big)^2 \\[1em] \Rightarrow \dfrac{5x}{2x - 6} = \dfrac{25}{9} \\[1em] \Rightarrow 5x \times 9 = 25(2x - 6) \\[1em] \Rightarrow 45x = 50x - 150 \\[1em] \Rightarrow 5x = 150 \\[1em] \Rightarrow x = 30. ⇒ ( 2 x − 6 5 x ) 2 = ( 3 5 ) 2 ⇒ 2 x − 6 5 x = 9 25 ⇒ 5 x × 9 = 25 ( 2 x − 6 ) ⇒ 45 x = 50 x − 150 ⇒ 5 x = 150 ⇒ x = 30.
Hence, the value of x is 30.
(v) Given,
a + x + a − x a + x − a − x = c d \dfrac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = \dfrac{c}{d} a + x − a − x a + x + a − x = d c
By componendo and dividendo,
⇒ a + x + a − x + a + x − a − x a + x + a − x − a + x + a − x = c + d c − d ⇒ 2 a + x 2 a − x = c + d c − d ⇒ a + x a − x = c + d c − d \Rightarrow \dfrac{\sqrt{a + x} + \sqrt{a - x} + \sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x} - \sqrt{a + x} + \sqrt{a - x}} = \dfrac{c + d}{c - d} \\[1em] \Rightarrow \dfrac{2\sqrt{a + x}}{2\sqrt{a - x}} = \dfrac{c + d}{c - d} \\[1em] \Rightarrow \dfrac{\sqrt{a + x}}{\sqrt{a - x}} = \dfrac{c + d}{c - d} ⇒ a + x + a − x − a + x + a − x a + x + a − x + a + x − a − x = c − d c + d ⇒ 2 a − x 2 a + x = c − d c + d ⇒ a − x a + x = c − d c + d
Squaring both sides,
⇒ a + x a − x = ( c + d c − d ) 2 ⇒ a + x a − x = c 2 + d 2 + 2 c d c 2 + d 2 − 2 c d \Rightarrow \dfrac{a + x}{a - x} = \Big(\dfrac{c + d}{c - d}\Big)^2 \\[1em] \Rightarrow \dfrac{a + x}{a - x} = \dfrac{c^2 + d^2 + 2cd}{c^2 + d^2 - 2cd} \\[1em] ⇒ a − x a + x = ( c − d c + d ) 2 ⇒ a − x a + x = c 2 + d 2 − 2 c d c 2 + d 2 + 2 c d
Again applying componendo and dividendo,
⇒ a + x + a − x a + x − a + x = c 2 + d 2 + 2 c d + c 2 + d 2 − 2 c d c 2 + d 2 + 2 c d − c 2 − d 2 + 2 c d ⇒ 2 a 2 x = 2 ( c 2 + d 2 ) 4 c d ⇒ a x = c 2 + d 2 2 c d ⇒ x = 2 a c d c 2 + d 2 \Rightarrow \dfrac{a + x + a - x}{a + x - a + x} = \dfrac{c^2 + d^2 + 2cd + c^2 + d^2 - 2cd}{c^2 + d^2 + 2cd - c^2 - d^2 + 2cd} \\[1em] \Rightarrow \dfrac{2a}{2x} = \dfrac{2(c^2 + d^2)}{4cd} \\[1em] \Rightarrow \dfrac{a}{x} = \dfrac{c^2 + d^2}{2cd} \\[1em] \Rightarrow x = \dfrac{2acd}{c^2 + d^2} ⇒ a + x − a + x a + x + a − x = c 2 + d 2 + 2 c d − c 2 − d 2 + 2 c d c 2 + d 2 + 2 c d + c 2 + d 2 − 2 c d ⇒ 2 x 2 a = 4 c d 2 ( c 2 + d 2 ) ⇒ x a = 2 c d c 2 + d 2 ⇒ x = c 2 + d 2 2 a c d
Hence, the value of x is 2 a c d c 2 + d 2 . \dfrac{2acd}{c^2 + d^2}. c 2 + d 2 2 a c d .
(vi) Given,
a + a 2 − 2 a x a − a 2 − 2 a x = b 1 . \dfrac{a + \sqrt{a^2 - 2ax}}{a - \sqrt{a^2 - 2ax}} = \dfrac{b}{1}. a − a 2 − 2 a x a + a 2 − 2 a x = 1 b .
By componendo and dividendo,
⇒ a + a 2 − 2 a x + a − a 2 − 2 a x a + a 2 − 2 a x − a + a 2 − 2 a x = b + 1 b − 1 ⇒ 2 a 2 a 2 − 2 a x = b + 1 b − 1 ⇒ a a 2 − 2 a x = b + 1 b − 1 \Rightarrow \dfrac{a + \sqrt{a^2 - 2ax} + a - \sqrt{a^2 - 2ax}}{a + \sqrt{a^2 - 2ax} - a + \sqrt{a^2 - 2ax}} = \dfrac{b + 1}{b - 1} \\[1em] \Rightarrow \dfrac{2a}{2\sqrt{a^2 - 2ax}} = \dfrac{b + 1}{b - 1} \\[1em] \Rightarrow \dfrac{a}{\sqrt{a^2 - 2ax}} = \dfrac{b + 1}{b - 1} \\[1em] ⇒ a + a 2 − 2 a x − a + a 2 − 2 a x a + a 2 − 2 a x + a − a 2 − 2 a x = b − 1 b + 1 ⇒ 2 a 2 − 2 a x 2 a = b − 1 b + 1 ⇒ a 2 − 2 a x a = b − 1 b + 1
Squaring both sides,
⇒ a 2 a 2 − 2 a x = ( b + 1 b − 1 ) 2 ⇒ a 2 a 2 − 2 a x = b 2 + 1 + 2 b b 2 + 1 − 2 b \Rightarrow \dfrac{a^2}{a^2 - 2ax} = \Big(\dfrac{b + 1}{b - 1}\Big)^2 \\[1em] \Rightarrow \dfrac{a^2}{a^2 - 2ax} = \dfrac{b^2 + 1 + 2b}{b^2 + 1 - 2b} \\[1em] ⇒ a 2 − 2 a x a 2 = ( b − 1 b + 1 ) 2 ⇒ a 2 − 2 a x a 2 = b 2 + 1 − 2 b b 2 + 1 + 2 b
Applying componendo and dividendo again,
⇒ a 2 + a 2 − 2 a x a 2 − a 2 + 2 a x = b 2 + 1 + 2 b + b 2 + 1 − 2 b b 2 + 1 + 2 b − b 2 − 1 + 2 b ⇒ 2 a ( a − x ) 2 a x = 2 ( b 2 + 1 ) 4 b ⇒ a − x x = b 2 + 1 2 b \Rightarrow \dfrac{a^2 + a^2 - 2ax}{a^2 - a^2 + 2ax} = \dfrac{b^2 + 1 + 2b + b^2 + 1 - 2b}{b^2 + 1 + 2b - b^2 - 1 + 2b} \\[1em] \Rightarrow \dfrac{2a(a - x)}{2ax} = \dfrac{2(b^2 + 1)}{4b} \\[1em] \Rightarrow \dfrac{a - x}{x} = \dfrac{b^2 + 1}{2b} ⇒ a 2 − a 2 + 2 a x a 2 + a 2 − 2 a x = b 2 + 1 + 2 b − b 2 − 1 + 2 b b 2 + 1 + 2 b + b 2 + 1 − 2 b ⇒ 2 a x 2 a ( a − x ) = 4 b 2 ( b 2 + 1 ) ⇒ x a − x = 2 b b 2 + 1
On cross-multiplication,
⇒ 2 b ( a − x ) = x ( b 2 + 1 ) ⇒ 2 a b − 2 b x = b 2 x + x ⇒ b 2 x + x + 2 b x = 2 a b ⇒ x ( b 2 + 1 + 2 b ) = 2 a b ⇒ x ( b + 1 ) 2 = 2 a b ⇒ x = 2 a b ( b + 1 ) 2 . \Rightarrow 2b(a - x) = x(b^2 + 1) \\[0.5em] \Rightarrow 2ab - 2bx = b^2x + x \\[0.5em] \Rightarrow b^2x + x + 2bx = 2ab \\[0.5em] \Rightarrow x(b^2 + 1 + 2b) = 2ab \\[0.5em] \Rightarrow x(b + 1)^2 = 2ab \\[0.5em] \Rightarrow x = \dfrac{2ab}{(b + 1)^2}. ⇒ 2 b ( a − x ) = x ( b 2 + 1 ) ⇒ 2 ab − 2 b x = b 2 x + x ⇒ b 2 x + x + 2 b x = 2 ab ⇒ x ( b 2 + 1 + 2 b ) = 2 ab ⇒ x ( b + 1 ) 2 = 2 ab ⇒ x = ( b + 1 ) 2 2 ab .
Hence, the value of x is 2 a b ( b + 1 ) 2 . \dfrac{2ab}{(b + 1)^2}. ( b + 1 ) 2 2 ab .
Using properties of proportion, solve for x. Given that x is positive.
(i) 3 x + 9 x 2 − 5 3 x − 9 x 2 − 5 = 5 (ii) 2 x + 4 x 2 − 1 2 x − 4 x 2 − 1 = 4 \begin{matrix} \text{(i)} & \dfrac{3x + \sqrt{9x^2 - 5}}{3x - \sqrt{9x^2 - 5}} = 5 \\[2em] \text{(ii)} & \dfrac{2x + \sqrt{4x^2 - 1}}{2x - \sqrt{4x^2 - 1}} = 4 \end{matrix} (i) (ii) 3 x − 9 x 2 − 5 3 x + 9 x 2 − 5 = 5 2 x − 4 x 2 − 1 2 x + 4 x 2 − 1 = 4
Answer
(i) Given,
3 x + 9 x 2 − 5 3 x − 9 x 2 − 5 = 5 1 . \dfrac{3x + \sqrt{9x^2 - 5}}{3x - \sqrt{9x^2 - 5}} = \dfrac{5}{1}. 3 x − 9 x 2 − 5 3 x + 9 x 2 − 5 = 1 5 .
By componendo and dividendo,
⇒ 3 x + 9 x 2 − 5 + 3 x − 9 x 2 − 5 3 x + 9 x 2 − 5 − 3 x + 9 x 2 − 5 = 5 + 1 5 − 1 ⇒ 6 x 2 9 x 2 − 5 = 6 4 ⇒ x 9 x 2 − 5 = 1 2 \Rightarrow \dfrac{3x + \sqrt{9x^2 - 5} + 3x - \sqrt{9x^2 - 5}}{3x + \sqrt{9x^2 - 5} - 3x + \sqrt{9x^2 - 5}} = \dfrac{5 + 1}{5 - 1} \\[1em] \Rightarrow \dfrac{6x}{2\sqrt{9x^2 - 5}} = \dfrac{6}{4} \\[1em] \Rightarrow \dfrac{x}{\sqrt{9x^2 - 5}} = \dfrac{1}{2} ⇒ 3 x + 9 x 2 − 5 − 3 x + 9 x 2 − 5 3 x + 9 x 2 − 5 + 3 x − 9 x 2 − 5 = 5 − 1 5 + 1 ⇒ 2 9 x 2 − 5 6 x = 4 6 ⇒ 9 x 2 − 5 x = 2 1
Squaring both sides we get,
x 2 9 x 2 − 5 = 1 4 ⇒ 4 x 2 = 9 x 2 − 5 ⇒ 5 x 2 − 5 = 0 ⇒ 5 ( x 2 − 1 ) = 0 ⇒ ( x + 1 ) ( x − 1 ) = 0 ⇒ x = 1 , − 1. \dfrac{x^2}{9x^2 - 5} = \dfrac{1}{4} \\[0.5em] \Rightarrow 4x^2 = 9x^2 - 5 \\[0.5em] \Rightarrow 5x^2 - 5 = 0 \\[0.5em] \Rightarrow 5(x^2 - 1) = 0 \\[0.5em] \Rightarrow (x + 1)(x - 1) = 0 \\[0.5em] \Rightarrow x = 1, -1. 9 x 2 − 5 x 2 = 4 1 ⇒ 4 x 2 = 9 x 2 − 5 ⇒ 5 x 2 − 5 = 0 ⇒ 5 ( x 2 − 1 ) = 0 ⇒ ( x + 1 ) ( x − 1 ) = 0 ⇒ x = 1 , − 1.
Since, x is positive, hence, x ≠ -1.
Hence, the required value of x is 1.
(ii) Given,
2 x + 4 x 2 − 1 2 x − 4 x 2 − 1 = 4 1 . \dfrac{2x + \sqrt{4x^2 - 1}}{2x - \sqrt{4x^2 - 1}} = \dfrac{4}{1}. 2 x − 4 x 2 − 1 2 x + 4 x 2 − 1 = 1 4 .
By componendo and dividendo,
⇒ 2 x + 4 x 2 − 1 + 2 x − 4 x 2 − 1 2 x + 4 x 2 − 1 − 2 x + 4 x 2 − 1 = 4 + 1 4 − 1 ⇒ 4 x 2 4 x 2 − 1 = 5 3 ⇒ 2 x 4 x 2 − 1 = 5 3 \Rightarrow \dfrac{2x + \sqrt{4x^2 - 1} + 2x - \sqrt{4x^2 - 1}}{2x + \sqrt{4x^2 - 1} - 2x + \sqrt{4x^2 - 1}} = \dfrac{4 + 1}{4 - 1} \\[1em] \Rightarrow \dfrac{4x}{2\sqrt{4x^2 - 1}} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{2x}{\sqrt{4x^2 - 1}} = \dfrac{5}{3} ⇒ 2 x + 4 x 2 − 1 − 2 x + 4 x 2 − 1 2 x + 4 x 2 − 1 + 2 x − 4 x 2 − 1 = 4 − 1 4 + 1 ⇒ 2 4 x 2 − 1 4 x = 3 5 ⇒ 4 x 2 − 1 2 x = 3 5
Squaring both sides we get,
4 x 2 4 x 2 − 1 = 25 9 ⇒ 4 x 2 × 9 = 25 ( 4 x 2 − 1 ) ⇒ 36 x 2 = 100 x 2 − 25 ⇒ 64 x 2 = 25 ⇒ x 2 = 25 64 ⇒ x = 25 64 ⇒ x = 5 8 or − 5 8 \dfrac{4x^2}{4x^2 - 1} = \dfrac{25}{9} \\[1em] \Rightarrow 4x^2 \times 9 = 25(4x^2 - 1) \\[1em] \Rightarrow 36x^2 = 100x^2 - 25 \\[1em] \Rightarrow 64x^2 = 25 \\[1em] \Rightarrow x^2 = \dfrac{25}{64} \\[1em] \Rightarrow x = \sqrt{\dfrac{25}{64}} \\[1em] \Rightarrow x = \dfrac{5}{8} \text{ or } -\dfrac{5}{8} \\[1em] 4 x 2 − 1 4 x 2 = 9 25 ⇒ 4 x 2 × 9 = 25 ( 4 x 2 − 1 ) ⇒ 36 x 2 = 100 x 2 − 25 ⇒ 64 x 2 = 25 ⇒ x 2 = 64 25 ⇒ x = 64 25 ⇒ x = 8 5 or − 8 5
Since, x is positive, hence, x ≠ − 5 8 . -\dfrac{5}{8}. − 8 5 .
Hence, the required value of x is 5 8 \dfrac{5}{8} 8 5
Solve : 1 + x + x 2 1 − x + x 2 = 62 ( 1 + x ) 63 ( 1 − x ) . \dfrac{1 + x + x^2}{1 - x + x^2} = \dfrac{62(1 + x)}{63(1 - x)}. 1 − x + x 2 1 + x + x 2 = 63 ( 1 − x ) 62 ( 1 + x ) .
Answer
Given,
1 + x + x 2 1 − x + x 2 = 62 ( 1 + x ) 63 ( 1 − x ) \dfrac{1 + x + x^2}{1 - x + x^2} = \dfrac{62(1 + x)}{63(1 - x)} 1 − x + x 2 1 + x + x 2 = 63 ( 1 − x ) 62 ( 1 + x )
⇒ ( 1 + x + x 2 ) ( 1 − x ) ( 1 − x + x 2 ) ( 1 + x ) = 62 63 ⇒ 1 + x + x 2 − x − x 2 − x 3 1 − x + x 2 + x − x 2 + x 3 = 62 63 ⇒ 1 − x + x − x 2 + x 2 − x 3 1 + x − x − x 2 + x 2 + x 3 = 62 63 ⇒ 1 − x 3 1 + x 3 = 62 63 \Rightarrow \dfrac{(1 + x + x^2)(1 - x)}{(1 - x + x^2)(1 + x)} = \dfrac{62}{63} \\[1em] \Rightarrow \dfrac{1 + x + x^2 -x -x^2 - x^3}{1 - x + x^2 + x - x^2 + x^3} = \dfrac{62}{63} \\[1em] \Rightarrow \dfrac{1 - \cancel{x} + \cancel{x} - \cancel{x^2} + \cancel{x^2} - x^3}{1 + \cancel{x} - \cancel{x} - \cancel{x^2} + \cancel{x^2} + x^3} = \dfrac{62}{63} \\[1em] \Rightarrow \dfrac{1 - x^3}{1 + x^3} = \dfrac{62}{63} ⇒ ( 1 − x + x 2 ) ( 1 + x ) ( 1 + x + x 2 ) ( 1 − x ) = 63 62 ⇒ 1 − x + x 2 + x − x 2 + x 3 1 + x + x 2 − x − x 2 − x 3 = 63 62 ⇒ 1 + x − x − x 2 + x 2 + x 3 1 − x + x − x 2 + x 2 − x 3 = 63 62 ⇒ 1 + x 3 1 − x 3 = 63 62
Again applying componendo and dividendo,
⇒ 1 − x 3 + 1 + x 3 1 − x 3 − 1 − x 3 = 62 + 63 62 − 63 ⇒ 2 − 2 x 3 = 125 − 1 ⇒ − 1 x 3 = − 125 ⇒ x 3 = 1 125 ⇒ x = 1 125 3 ⇒ x = 1 5 . \Rightarrow \dfrac{1 - x^3 + 1 + x^3}{1 - x^3 - 1 -x^3} = \dfrac{62 + 63}{62 - 63} \\[1em] \Rightarrow \dfrac{2}{-2x^3} = \dfrac{125}{-1} \\[1em] \Rightarrow -\dfrac{1}{x^3} = -125 \\[1em] \Rightarrow x^3 = \dfrac{1}{125} \\[1em] \Rightarrow x = \dfrac{1}{\sqrt[3]{125}} \\[1em] \Rightarrow x = \dfrac{1}{5}. ⇒ 1 − x 3 − 1 − x 3 1 − x 3 + 1 + x 3 = 62 − 63 62 + 63 ⇒ − 2 x 3 2 = − 1 125 ⇒ − x 3 1 = − 125 ⇒ x 3 = 125 1 ⇒ x = 3 125 1 ⇒ x = 5 1 .
Hence, the required value is 1 5 . \dfrac{1}{5}. 5 1 .
Solve for x : 16 ( a − x a + x ) 3 = a + x a − x . 16\Big(\dfrac{a - x}{a + x}\Big)^3 =\dfrac{a + x}{a - x}. 16 ( a + x a − x ) 3 = a − x a + x .
Answer
Given,
16 ( a − x a + x ) 3 = a + x a − x . 16\Big(\dfrac{a - x}{a + x}\Big)^3 =\dfrac{a + x}{a - x}. 16 ( a + x a − x ) 3 = a − x a + x .
⇒ ( a − x ) 3 ( a + x ) 3 × ( a − x ) ( a + x ) = 1 16 ⇒ ( a − x a + x ) 4 = ( 1 2 ) 4 or ( − 1 2 ) 4 ⇒ a − x a + x = 1 2 or − 1 2 \Rightarrow \dfrac{(a - x)^3}{(a + x)^3} \times \dfrac{(a - x)}{(a + x)} = \dfrac{1}{16} \\[1em] \Rightarrow \Big(\dfrac{a - x}{a + x}\Big)^4 = \Big(\dfrac{1}{2}\Big)^4 \text{ or } \Big(-\dfrac{1}{2}\Big)^4 \\[1em] \Rightarrow \dfrac{a - x}{a + x} = \dfrac{1}{2} \text{ or } -\dfrac{1}{2} ⇒ ( a + x ) 3 ( a − x ) 3 × ( a + x ) ( a − x ) = 16 1 ⇒ ( a + x a − x ) 4 = ( 2 1 ) 4 or ( − 2 1 ) 4 ⇒ a + x a − x = 2 1 or − 2 1
First Solving,
a − x a + x = 1 2 \dfrac{a - x}{a + x} = \dfrac{1}{2} a + x a − x = 2 1
By componendo and dividendo,
⇒ a − x + a + x a − x − a − x = 1 + 2 1 − 2 ⇒ − 2 a 2 x = − 3 ⇒ a x = 3 ⇒ x = a 3 . \Rightarrow \dfrac{a - x + a + x}{a - x - a - x} = \dfrac{1 + 2}{1 - 2} \\[1em] \Rightarrow -\dfrac{2a}{2x} = -3 \\[1em] \Rightarrow \dfrac{a}{x} = 3 \\[1em] \Rightarrow x = \dfrac{a}{3}. ⇒ a − x − a − x a − x + a + x = 1 − 2 1 + 2 ⇒ − 2 x 2 a = − 3 ⇒ x a = 3 ⇒ x = 3 a .
Now Solving,
a − x a + x = − 1 2 \dfrac{a - x}{a + x} = -\dfrac{1}{2} a + x a − x = − 2 1
By componendo and dividendo,
⇒ a − x + a + x a − x − a − x = 1 − 2 1 + 2 ⇒ − 2 a 2 x = − 1 3 ⇒ a x = 1 3 ⇒ x = 3 a . \Rightarrow \dfrac{a - x + a + x}{a - x - a - x} = \dfrac{1 - 2}{1 + 2}\\[1em] \Rightarrow -\dfrac{2a}{2x} = -\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{a}{x} = \dfrac{1}{3} \\[1em] \Rightarrow x = 3a. \\[1em] ⇒ a − x − a − x a − x + a + x = 1 + 2 1 − 2 ⇒ − 2 x 2 a = − 3 1 ⇒ x a = 3 1 ⇒ x = 3 a .
Hence, the value of x is a 3 \dfrac{a}{3} 3 a
If x = a + 1 + a − 1 a + 1 − a − 1 , \dfrac{\sqrt{a + 1} + \sqrt{a - 1}}{\sqrt{a + 1} - \sqrt{a - 1}}, a + 1 − a − 1 a + 1 + a − 1 ,
x2 - 2ax + 1 = 0.
Answer
Given,
x 1 = a + 1 + a − 1 a + 1 − a − 1 . \dfrac{x}{1} = \dfrac{\sqrt{a + 1} + \sqrt{a - 1}}{\sqrt{a + 1} - \sqrt{a - 1}}. 1 x = a + 1 − a − 1 a + 1 + a − 1 .
Applying componendo and dividendo,
⇒ x + 1 x − 1 = a + 1 + a − 1 + a + 1 − a − 1 a + 1 + a − 1 − a + 1 + a − 1 ⇒ x + 1 x − 1 = 2 a + 1 2 a − 1 ⇒ x + 1 x − 1 = a + 1 a − 1 \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{a + 1} + \sqrt{a - 1} + \sqrt{a + 1} - \sqrt{a - 1}}{\sqrt{a + 1} + \sqrt{a - 1} - \sqrt{a + 1} + \sqrt{a - 1}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{a + 1}}{2\sqrt{a - 1}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{a + 1}}{\sqrt{a - 1}} ⇒ x − 1 x + 1 = a + 1 + a − 1 − a + 1 + a − 1 a + 1 + a − 1 + a + 1 − a − 1 ⇒ x − 1 x + 1 = 2 a − 1 2 a + 1 ⇒ x − 1 x + 1 = a − 1 a + 1
Squaring both sides we get,
⇒ ( x + 1 x − 1 ) 2 = ( a + 1 a − 1 ) 2 ⇒ x 2 + 1 + 2 x x 2 + 1 − 2 x = a + 1 a − 1 \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^2 = \Big(\dfrac{\sqrt{a + 1}}{\sqrt{a - 1}}\Big)^2 \\[1em] \Rightarrow \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \dfrac{a + 1}{a - 1} ⇒ ( x − 1 x + 1 ) 2 = ( a − 1 a + 1 ) 2 ⇒ x 2 + 1 − 2 x x 2 + 1 + 2 x = a − 1 a + 1
Again applying componendo and dividendo,
⇒ x 2 + 1 + 2 x + x 2 + 1 − 2 x x 2 + 1 + 2 x − x 2 − 1 + 2 x = a + 1 + a − 1 a + 1 − a + 1 ⇒ 2 ( x 2 + 1 ) 4 x = 2 a 2 ⇒ x 2 + 1 2 x = a ⇒ x 2 + 1 = 2 a x ⇒ x 2 − 2 a x + 1 = 0. \Rightarrow \dfrac{x^2 + 1 + \cancel{2x} + x^2 + 1 - \cancel{2x}}{\cancel{x^2} + \cancel{1} + 2x - \cancel{x^2} - \cancel{1} + 2x} = \dfrac{a + \cancel{1} + a - \cancel{1}}{\cancel{a} + 1 - \cancel{a} + 1} \\[1em] \Rightarrow \dfrac{2(x^2 + 1)}{4x} = \dfrac{2a}{2} \\[1em] \Rightarrow \dfrac{x^2 + 1}{2x} = a \\[1em] \Rightarrow x^2 + 1 = 2ax \\[1em] \Rightarrow x^2 - 2ax + 1 = 0. ⇒ x 2 + 1 + 2 x − x 2 − 1 + 2 x x 2 + 1 + 2 x + x 2 + 1 − 2 x = a + 1 − a + 1 a + 1 + a − 1 ⇒ 4 x 2 ( x 2 + 1 ) = 2 2 a ⇒ 2 x x 2 + 1 = a ⇒ x 2 + 1 = 2 a x ⇒ x 2 − 2 a x + 1 = 0.
Hence proved that, x2 - 2ax + 1 = 0.
If x = 2 a + 1 + 2 a − 1 2 a + 1 − 2 a − 1 \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}} 2 a + 1 − 2 a − 1 2 a + 1 + 2 a − 1 2 - 4ax + 1 = 0.
Answer
Given; x = 2 a + 1 + 2 a − 1 2 a + 1 − 2 a − 1 \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}} 2 a + 1 − 2 a − 1 2 a + 1 + 2 a − 1
Applying componendo and dividendo on both sides we get :
⇒ x + 1 x − 1 = 2 a + 1 + 2 a − 1 + 2 a + 1 − 2 a − 1 2 a + 1 + 2 a − 1 − ( 2 a + 1 − 2 a − 1 ) ⇒ x + 1 x − 1 = 2 2 a + 1 2 2 a − 1 ⇒ x + 1 x − 1 = 2 a + 1 2 a − 1 Squaring both sides we get : ⇒ ( x + 1 ) 2 ( x − 1 ) 2 = 2 a + 1 2 a − 1 ⇒ x 2 + 1 + 2 x x 2 + 1 − 2 x = 2 a + 1 2 a − 1 Applying componendo and dividendo on both sides we get : ⇒ x 2 + 1 + 2 x + x 2 + 1 − 2 x x 2 + 1 + 2 x − ( x 2 + 1 − 2 x ) = 2 a + 1 + 2 a − 1 2 a + 1 − ( 2 a − 1 ) ⇒ 2 x 2 + 2 4 x = 4 a 2 ⇒ x 2 + 1 2 x = 2 a ⇒ x 2 + 1 = 4 a x ⇒ x 2 + 1 − 4 a x = 0 \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1} + \sqrt{2a + 1} - \sqrt{2a - 1}}{\sqrt{2a + 1} + \sqrt{2a - 1} - (\sqrt{2a + 1} - \sqrt{2a - 1})}\\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{2a + 1}}{2\sqrt{2a - 1}}\\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 1}}{\sqrt{2a - 1}}\\[1em] \text{Squaring both sides we get :}\\[1em] \Rightarrow \dfrac{(x + 1)^2}{(x - 1)^2} = \dfrac{2a + 1}{2a - 1}\\[1em] \Rightarrow \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \dfrac{2a + 1}{2a - 1}\\[1em] \text{Applying componendo and dividendo on both sides we get :}\\[1em] \Rightarrow \dfrac{x^2 + 1 + 2x + x^2 + 1 - 2x}{x^2 + 1 + 2x - (x^2 + 1 - 2x)} = \dfrac{2a + 1 + 2a - 1}{2a + 1 - (2a - 1)}\\[1em] \Rightarrow \dfrac{2x^2 + 2}{4x} = \dfrac{4a}{2}\\[1em] \Rightarrow \dfrac{x^2 + 1}{2x} = 2a\\[1em] \Rightarrow x^2 + 1 = 4ax\\[1em] \Rightarrow x^2 + 1 - 4ax = 0 ⇒ x − 1 x + 1 = 2 a + 1 + 2 a − 1 − ( 2 a + 1 − 2 a − 1 ) 2 a + 1 + 2 a − 1 + 2 a + 1 − 2 a − 1 ⇒ x − 1 x + 1 = 2 2 a − 1 2 2 a + 1 ⇒ x − 1 x + 1 = 2 a − 1 2 a + 1 Squaring both sides we get : ⇒ ( x − 1 ) 2 ( x + 1 ) 2 = 2 a − 1 2 a + 1 ⇒ x 2 + 1 − 2 x x 2 + 1 + 2 x = 2 a − 1 2 a + 1 Applying componendo and dividendo on both sides we get : ⇒ x 2 + 1 + 2 x − ( x 2 + 1 − 2 x ) x 2 + 1 + 2 x + x 2 + 1 − 2 x = 2 a + 1 − ( 2 a − 1 ) 2 a + 1 + 2 a − 1 ⇒ 4 x 2 x 2 + 2 = 2 4 a ⇒ 2 x x 2 + 1 = 2 a ⇒ x 2 + 1 = 4 a x ⇒ x 2 + 1 − 4 a x = 0
Hence, proved that x2 - 4ax + 1 = 0.
Given x = a 2 + b 2 + a 2 − b 2 a 2 + b 2 − a 2 − b 2 , \dfrac{\sqrt{a^{2} + b^2} + \sqrt{a^2 - b^2}}{\sqrt{a^2 + b^2} - \sqrt{a^2 - b^2}}, a 2 + b 2 − a 2 − b 2 a 2 + b 2 + a 2 − b 2 , 2 = 2 a 2 x x 2 + 1 . \dfrac{2a^2x}{x^2 + 1}. x 2 + 1 2 a 2 x .
Answer
Given,
x = a 2 + b 2 + a 2 − b 2 a 2 + b 2 − a 2 − b 2 x = \dfrac{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2}}{\sqrt{a^2 + b^2} - \sqrt{a^2 - b^2}} x = a 2 + b 2 − a 2 − b 2 a 2 + b 2 + a 2 − b 2
By componendo and dividendo,
⇒ x + 1 x − 1 = a 2 + b 2 + a 2 − b 2 + a 2 + b 2 − a 2 − b 2 a 2 + b 2 + a 2 − b 2 − a 2 + b 2 + a 2 − b 2 ⇒ x + 1 x − 1 = 2 a 2 + b 2 2 a 2 − b 2 ⇒ x + 1 x − 1 = a 2 + b 2 a 2 − b 2 \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2} + \sqrt{a^2 + b^2} - \sqrt{a^2 - b^2}}{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2} - \sqrt{a^2 + b^2} + \sqrt{a^2 - b^2}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{a^2 + b^2}}{2\sqrt{a^2 - b^2}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{a^2 + b^2}}{\sqrt{a^2 - b^2}} ⇒ x − 1 x + 1 = a 2 + b 2 + a 2 − b 2 − a 2 + b 2 + a 2 − b 2 a 2 + b 2 + a 2 − b 2 + a 2 + b 2 − a 2 − b 2 ⇒ x − 1 x + 1 = 2 a 2 − b 2 2 a 2 + b 2 ⇒ x − 1 x + 1 = a 2 − b 2 a 2 + b 2
On squaring both sides,
⇒ ( x + 1 x − 1 ) 2 = a 2 + b 2 a 2 − b 2 ⇒ x 2 + 1 + 2 x x 2 + 1 − 2 x = a 2 + b 2 a 2 − b 2 \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^2 = \dfrac{a^2 + b^2}{a^2 - b^2} \\[1em] \Rightarrow \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \dfrac{a^2 + b^2}{a^2 - b^2} \\[1em] ⇒ ( x − 1 x + 1 ) 2 = a 2 − b 2 a 2 + b 2 ⇒ x 2 + 1 − 2 x x 2 + 1 + 2 x = a 2 − b 2 a 2 + b 2
Again applying componendo and dividendo,
⇒ x 2 + 1 + 2 x + x 2 + 1 − 2 x x 2 + 1 + 2 x − x 2 − 1 + 2 x = a 2 + b 2 + a 2 − b 2 a 2 + b 2 − a 2 + b 2 ⇒ 2 x 2 + 2 4 x = 2 a 2 2 b 2 ⇒ x 2 + 1 2 x = a 2 b 2 ⇒ b 2 = 2 a 2 x x 2 + 1 . \Rightarrow \dfrac{x^2 + 1 + \cancel{2x} + x^2 + 1 - \cancel{2x}}{\cancel{x^2} + \cancel{1} + 2x - \cancel{x^2} - \cancel{1} + 2x} = \dfrac{a^2 + \cancel{b^2} + a^2 - \cancel{b^2}}{\cancel{a^2} + b^2 - \cancel{a^2} + b^2} \\[1em] \Rightarrow \dfrac{2x^2 + 2}{4x} = \dfrac{2a^2}{2b^2} \\[1em] \Rightarrow \dfrac{x^2 + 1}{2x} = \dfrac{a^2}{b^2}\\[1em] \Rightarrow b^2 = \dfrac{2a^2x}{x^2 + 1}. ⇒ x 2 + 1 + 2 x − x 2 − 1 + 2 x x 2 + 1 + 2 x + x 2 + 1 − 2 x = a 2 + b 2 − a 2 + b 2 a 2 + b 2 + a 2 − b 2 ⇒ 4 x 2 x 2 + 2 = 2 b 2 2 a 2 ⇒ 2 x x 2 + 1 = b 2 a 2 ⇒ b 2 = x 2 + 1 2 a 2 x .
Hence, proved that b 2 = 2 a 2 x x 2 + 1 . b^2 = \dfrac{2a^2x}{x^2 + 1}. b 2 = x 2 + 1 2 a 2 x .
Given that a 3 + 3 a b 2 b 3 + 3 a 2 b = 63 62 . \dfrac{a^3 + 3ab^2}{b^3 + 3a^2b} = \dfrac{63}{62}. b 3 + 3 a 2 b a 3 + 3 a b 2 = 62 63 .
Answer
Given,
a 3 + 3 a b 2 b 3 + 3 a 2 b = 63 62 \dfrac{a^3 + 3ab^2}{b^3 + 3a^2b} = \dfrac{63}{62} b 3 + 3 a 2 b a 3 + 3 a b 2 = 62 63
By componendo and dividendo,
⇒ a 3 + 3 a b 2 + b 3 + 3 a 2 b a 3 + 3 a b 2 − b 3 − 3 a 2 b = 63 + 62 63 − 62 ⇒ ( a + b a − b ) 3 = 125 ⇒ ( a + b a − b ) 3 = ( 5 ) 3 ⇒ a + b a − b = 5 ⇒ a + b = 5 a − 5 b ⇒ 5 a − a = b + 5 b ⇒ 4 a = 6 b ⇒ a b = 6 4 = 2 3 ⇒ a : b = 3 : 2. \Rightarrow \dfrac{a^3 + 3ab^2 + b^3 + 3a^2b}{a^3 + 3ab^2 - b^3 - 3a^2b} = \dfrac{63 + 62}{63 - 62} \\[1em] \Rightarrow \Big(\dfrac{a + b}{a - b}\Big)^3 = 125 \\[1em] \Rightarrow \Big(\dfrac{a + b}{a - b}\Big)^3 = (5)^3 \\[1em] \Rightarrow \dfrac{a + b}{a - b} = 5 \\[1em] \Rightarrow a + b = 5a - 5b \\[1em] \Rightarrow 5a - a = b + 5b \\[1em] \Rightarrow 4a = 6b \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{6}{4} = \dfrac{2}{3} \\[1em] \Rightarrow a : b = 3 : 2. ⇒ a 3 + 3 a b 2 − b 3 − 3 a 2 b a 3 + 3 a b 2 + b 3 + 3 a 2 b = 63 − 62 63 + 62 ⇒ ( a − b a + b ) 3 = 125 ⇒ ( a − b a + b ) 3 = ( 5 ) 3 ⇒ a − b a + b = 5 ⇒ a + b = 5 a − 5 b ⇒ 5 a − a = b + 5 b ⇒ 4 a = 6 b ⇒ b a = 4 6 = 3 2 ⇒ a : b = 3 : 2.
Hence, the value of a : b = 3 : 2.
Given x 3 + 12 x 6 x 2 + 8 = y 3 + 27 y 9 y 2 + 27 . \dfrac{x^3 + 12x}{6x^2 + 8} = \dfrac{y^3 + 27y}{9y^2 + 27}. 6 x 2 + 8 x 3 + 12 x = 9 y 2 + 27 y 3 + 27 y .
Answer
Given,
x 3 + 12 x 6 x 2 + 8 = y 3 + 27 y 9 y 2 + 27 \dfrac{x^3 + 12x}{6x^2 + 8} = \dfrac{y^3 + 27y}{9y^2 + 27} 6 x 2 + 8 x 3 + 12 x = 9 y 2 + 27 y 3 + 27 y
By componendo and dividendo,
⇒ x 3 + 12 x + 6 x 2 + 8 x 3 + 12 x − 6 x 2 − 8 = y 3 + 27 y + 9 y 2 + 27 y 3 + 27 y − 9 y 2 − 27 ⇒ x 3 + ( 3 × x × 2 2 ) + ( 3 × x 2 × 2 ) + 2 3 x 3 + ( 3 × x × 2 2 ) − ( 3 × x 2 × 2 ) − 2 3 = y 3 + ( 3 × y × 3 2 ) + ( 3 × y 2 × 3 ) + 3 3 y 3 + ( 3 × y × 3 2 ) − ( 3 × y 2 × 3 ) − 3 3 ⇒ ( x + 2 x − 2 ) 3 = ( y + 3 y − 3 ) 3 ⇒ x + 2 x − 2 = y + 3 y − 3 \Rightarrow \dfrac{x^3 + 12x + 6x^2 + 8}{x^3 + 12x - 6x^2 - 8} = \dfrac{y^3 + 27y + 9y^2 + 27}{y^3 + 27y - 9y^2 - 27} \\[1em] \Rightarrow \dfrac{x^3 + (3 \times x \times 2^2 ) + (3 \times x^2 \times 2 ) + 2^3}{x^3 + (3 \times x \times 2^2 ) - (3 \times x^2 \times 2 ) - 2^3} \\[1em] = \dfrac{y^3 + (3 \times y \times 3^2) + (3 \times y^2 \times 3) + 3^3}{y^3 + (3 \times y \times 3^2) - (3 \times y^2 \times 3) - 3^3} \\[1em] \Rightarrow \Big(\dfrac{x + 2}{x - 2}\Big)^3 = \Big(\dfrac{y + 3}{y - 3}\Big)^3 \\[1em] \Rightarrow \dfrac{x + 2}{x - 2} = \dfrac{y + 3}{y - 3} ⇒ x 3 + 12 x − 6 x 2 − 8 x 3 + 12 x + 6 x 2 + 8 = y 3 + 27 y − 9 y 2 − 27 y 3 + 27 y + 9 y 2 + 27 ⇒ x 3 + ( 3 × x × 2 2 ) − ( 3 × x 2 × 2 ) − 2 3 x 3 + ( 3 × x × 2 2 ) + ( 3 × x 2 × 2 ) + 2 3 = y 3 + ( 3 × y × 3 2 ) − ( 3 × y 2 × 3 ) − 3 3 y 3 + ( 3 × y × 3 2 ) + ( 3 × y 2 × 3 ) + 3 3 ⇒ ( x − 2 x + 2 ) 3 = ( y − 3 y + 3 ) 3 ⇒ x − 2 x + 2 = y − 3 y + 3
Again applying componendo and dividendo,
⇒ x + 2 + x − 2 x + 2 − x + 2 = y + 3 + y − 3 y + 3 − y + 3 ⇒ 2 x 4 = 2 y 6 ⇒ x y = 2 6 × 4 2 ⇒ x y = 2 3 ⇒ x : y = 2 : 3. \Rightarrow \dfrac{x + \cancel{2} + x - \cancel{2}}{\cancel{x} + 2 - \cancel{x} + 2} = \dfrac{y + \cancel{3} + y - \cancel{3}}{\cancel{y} + 3 - \cancel{y} + 3} \\[1em] \Rightarrow \dfrac{2x}{4} = \dfrac{2y}{6} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{2}{6} \times \dfrac{4}{2} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{2}{3} \\[1em] \Rightarrow x : y = 2 : 3. ⇒ x + 2 − x + 2 x + 2 + x − 2 = y + 3 − y + 3 y + 3 + y − 3 ⇒ 4 2 x = 6 2 y ⇒ y x = 6 2 × 2 4 ⇒ y x = 3 2 ⇒ x : y = 2 : 3.
Hence, the value of ratio x : y is 2 : 3.
Using the properties of proportion, solve the following equation for x;
given x 3 + 3 x 3 x 2 + 1 = 341 91 . \dfrac{x^3 + 3x}{3x^2 + 1} = \dfrac{341}{91}. 3 x 2 + 1 x 3 + 3 x = 91 341 .
Answer
Given,
x 3 + 3 x 3 x 2 + 1 = 341 91 \dfrac{x^3 + 3x}{3x^2 + 1} = \dfrac{341}{91} 3 x 2 + 1 x 3 + 3 x = 91 341
Applying componendo and dividendo,
⇒ x 3 + 3 x + 3 x 2 + 1 x 3 + 3 x − 3 x 2 − 1 = 341 + 91 341 − 91 ⇒ ( x + 1 ) 3 ( x − 1 ) 3 = 432 250 ⇒ ( x + 1 ) 3 ( x − 1 ) 3 = 216 125 ⇒ ( x + 1 x − 1 ) 3 = ( 6 5 ) 3 ⇒ x + 1 x − 1 = 6 5 \Rightarrow \dfrac{x^3 + 3x + 3x^2 + 1}{x^3 + 3x - 3x^2 - 1} = \dfrac{341 + 91}{341 - 91} \\[1em] \Rightarrow \dfrac{(x + 1)^3}{(x - 1)^3} = \dfrac{432}{250} \\[1em] \Rightarrow \dfrac{(x + 1)^3}{(x - 1)^3} = \dfrac{216}{125} \\[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3 = \Big(\dfrac{6}{5}\Big)^3 \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{6}{5} ⇒ x 3 + 3 x − 3 x 2 − 1 x 3 + 3 x + 3 x 2 + 1 = 341 − 91 341 + 91 ⇒ ( x − 1 ) 3 ( x + 1 ) 3 = 250 432 ⇒ ( x − 1 ) 3 ( x + 1 ) 3 = 125 216 ⇒ ( x − 1 x + 1 ) 3 = ( 5 6 ) 3 ⇒ x − 1 x + 1 = 5 6
Again applying componendo and dividendo,
⇒ x + 1 + x − 1 x + 1 − x + 1 = 6 + 5 6 − 5 ⇒ 2 x 2 = 11 1 ⇒ x = 11. \Rightarrow \dfrac{x + \cancel{1} + x - \cancel{1}}{\cancel{x} + 1 - \cancel{x} + 1} = \dfrac{6 + 5}{6 - 5} \\[1em] \Rightarrow \dfrac{2x}{2} = \dfrac{11}{1} \\[1em] \Rightarrow x = 11. ⇒ x + 1 − x + 1 x + 1 + x − 1 = 6 − 5 6 + 5 ⇒ 2 2 x = 1 11 ⇒ x = 11.
Hence, the value of x is 11.
If x + y a x + b y = y + z a y + b z = z + x a z + b x \dfrac{x + y}{ax + by} = \dfrac{y + z}{ay + bz} = \dfrac{z + x}{az + bx} a x + b y x + y = a y + b z y + z = a z + b x z + x 2 a + b , \dfrac{2}{a + b}, a + b 2 ,
Answer
We know that if a b = c d = e f \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} b a = d c = f e
= a + c + e b + d + f = sum of antecedents sum of consequents = \dfrac{a + c + e}{b + d + f} = \dfrac{\text{sum of antecedents}}{\text{sum of consequents}} = b + d + f a + c + e = sum of consequents sum of antecedents
∴ x + y a x + b y = y + z a y + b z = z + x a z + b x = x + y + y + z + z + x a x + b y + a y + b z + a z + b x = 2 ( x + y + z ) a ( x + y + z ) + b ( x + y + z ) = 2 ( x + y + z ) ( a + b ) ( x + y + z ) = 2 a + b . \therefore \dfrac{x + y}{ax + by} = \dfrac{y + z}{ay + bz} = \dfrac{z + x}{az + bx} \\[1em] = \dfrac{x + y + y + z + z + x}{ax + by + ay + bz + az + bx} \\[1em] = \dfrac{2(x + y + z)}{a(x + y + z) + b(x + y + z)} \\[1em] = \dfrac{2(x + y + z)}{(a + b)(x + y + z)} \\[1em] = \dfrac{2}{a + b}. ∴ a x + b y x + y = a y + b z y + z = a z + b x z + x = a x + b y + a y + b z + a z + b x x + y + y + z + z + x = a ( x + y + z ) + b ( x + y + z ) 2 ( x + y + z ) = ( a + b ) ( x + y + z ) 2 ( x + y + z ) = a + b 2 .
Hence, proved that,
x + y a x + b y = y + z a y + b z = z + x a z + b x = 2 a + b . \dfrac{x + y}{ax + by} = \dfrac{y + z}{ay + bz} = \dfrac{z + x}{az + bx} = \dfrac{2}{a + b}. a x + b y x + y = a y + b z y + z = a z + b x z + x = a + b 2 .
