Algebraic Expressions and Identities — Exercise 10.5

Using suitable identities, find the following product:

$(3x + 5) (3x + 5)$

Answer

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$,

$\Rightarrow (3x + 5)(3x + 5)\\[1em] \Rightarrow (3x + 5)^2\\[1em] \Rightarrow (3x)^2 + 2 \times 3x \times 5 + (5)^2\\[1em] \Rightarrow 9x^2 + 30x + 25$

Hence, $\bm{(3x + 5)(3x + 5) = 9x^2 + 30x + 25}$.

Using suitable identities, find the following product:

$(9y - 5)(9y - 5)$

Answer

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

$\Rightarrow (9y - 5)(9y - 5)\\[1em] \Rightarrow (9y - 5)^2\\[1em] \Rightarrow (9y)^2 - 2 \times 9y \times 5 + (5)^2\\[1em] \Rightarrow 81y^2 - 90y + 25$

Hence, $\bm{(9y - 5)(9y - 5) = 81y^2 - 90y + 25}$.

Using suitable identities, find the following product:

$(4x + 11y)(4x - 11y)$

Answer

Using the identity $(a + b)(a - b) = a^2 - b^2$,

$\Rightarrow (4x + 11y)(4x - 11y)\\[1em] \Rightarrow (4x)^2 - (11y)^2\\[1em] \Rightarrow 16x^2 - 121y^2$

Hence, $\bm{(4x + 11y)(4x - 11y) = 16x^2 - 121y^2}$.

Using suitable identities, find the following product:

$\left(\dfrac{3}{2}m + \dfrac{2}{3}n\right)\left(\dfrac{3}{2}m - \dfrac{2}{3}n\right)$

Answer

Using the identity $(a + b)(a - b) = a^2 - b^2$,

$\Rightarrow \Big(\dfrac{3}{2}m + \dfrac{2}{3}n\Big)\Big(\dfrac{3}{2}m - \dfrac{2}{3}n\Big)\\[1em] \Rightarrow \Big(\dfrac{3}{2}m\Big)^2 - \Big(\dfrac{2}{3}n\Big)^2\\[1em] \Rightarrow \dfrac{9}{4}m^2 - \dfrac{4}{9}n^2$

Hence, $\bm{\Big(\dfrac{3}{2}m + \dfrac{2}{3}n\Big)\Big(\dfrac{3}{2}m - \dfrac{2}{3}n\Big) = \dfrac{9}{4}m^2 - \dfrac{4}{9}n^2}$.

Using suitable identities, find the following product:

$\left(\dfrac{2}{a} + \dfrac{5}{b}\right)\left(\dfrac{2}{a} + \dfrac{5}{b}\right)$

Answer

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$,

$\Rightarrow \Big(\dfrac{2}{a} + \dfrac{5}{b}\Big)\Big(\dfrac{2}{a} + \dfrac{5}{b}\Big)\\[1em] \Rightarrow \Big(\dfrac{2}{a} + \dfrac{5}{b}\Big)^2\\[1em] \Rightarrow \Big(\dfrac{2}{a}\Big)^2 + 2 \times \dfrac{2}{a} \times \dfrac{5}{b} + \Big(\dfrac{5}{b}\Big)^2\\[1em] \Rightarrow \dfrac{4}{a^2} + \dfrac{20}{ab} + \dfrac{25}{b^2}$

Hence, $\bm{\Big(\dfrac{2}{a} + \dfrac{5}{b}\Big)\Big(\dfrac{2}{a} + \dfrac{5}{b}\Big) = \dfrac{4}{a^2} + \dfrac{20}{ab} + \dfrac{25}{b^2}}$.

Using suitable identities, find the following product:

$\left(\dfrac{p^2}{2} + \dfrac{2}{q^2}\right)\left(\dfrac{p^2}{2} - \dfrac{2}{q^2}\right)$

Answer

Using the identity $(a + b)(a - b) = a^2 - b^2$,

$\Rightarrow \Big(\dfrac{p^2}{2} + \dfrac{2}{q^2}\Big)\Big(\dfrac{p^2}{2} - \dfrac{2}{q^2}\Big)\\[1em] \Rightarrow \Big(\dfrac{p^2}{2}\Big)^2 - \Big(\dfrac{2}{q^2}\Big)^2\\[1em] \Rightarrow \dfrac{p^4}{4} - \dfrac{4}{q^4}$

Hence, $\bm{\Big(\dfrac{p^2}{2} + \dfrac{2}{q^2}\Big)\Big(\dfrac{p^2}{2} - \dfrac{2}{q^2}\Big) = \dfrac{p^4}{4} - \dfrac{4}{q^4}}$.

Using the identities, evaluate the following:

81²

Answer

⇒ 81²

⇒ (80 + 1)²

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$

⇒ (80)² + 2 × 80 × 1 + (1)²

⇒ 6400 + 160 + 1

⇒ 6561

Hence, 81² = 6561.

Using the identities, evaluate the following:

97²

Answer

⇒ 97²

⇒ (100 - 3)²

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$

⇒ (100)² - 2 × 100 × 3 + (3)²

⇒ 10000 - 600 + 9

⇒ 9409

Hence, 97² = 9409.

Using the identities, evaluate the following:

105²

Answer

⇒ 105²

⇒ (100 + 5)²

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$

⇒ (100)² + 2 × 100 × 5 + (5)²

⇒ 10000 + 1000 + 25

⇒ 11025

Hence, 105² = 11025.

Using the identities, evaluate the following:

997²

Answer

⇒ 997²

⇒ (1000 - 3)²

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$

⇒ (1000)² - 2 × 1000 × 3 + (3)²

⇒ 1000000 - 6000 + 9

⇒ 994009

Hence, 997² = 994009.

Using the identities, evaluate the following:

6.1²

Answer

⇒ 6.1²

⇒ (6 + 0.1)²

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$

⇒ (6)² + 2 × 6 × 0.1 + (0.1)²

⇒ 36 + 1.2 + 0.01

⇒ 37.21

Hence, 6.1² = 37.21.

Using the identities, evaluate the following:

496 × 504

Answer

⇒ 496 × 504

⇒ (500 - 4)(500 + 4)

Using the identity $(a - b)(a + b) = a^2 - b^2$

⇒ (500)² - (4)²

⇒ 250000 - 16

⇒ 249984

Hence, 496 × 504 = 249984.

Using the identities, evaluate the following:

20.5 × 19.5

Answer

⇒ 20.5 × 19.5

⇒ (20 + 0.5)(20 - 0.5)

Using the identity $(a + b)(a - b) = a^2 - b^2$

⇒ (20)² - (0.5)²

⇒ 400 - 0.25

⇒ 399.75

Hence, 20.5 × 19.5 = 399.75.

Using the identities, evaluate the following:

9.6²

Answer

⇒ 9.6²

⇒ (10 - 0.4)²

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$

⇒ (10)² - 2 × 10 × 0.4 + (0.4)²

⇒ 100 - 8 + 0.16

⇒ 92.16

Hence, 9.6² = 92.16.

Find the following square, using the identities:

$(pq + 5r)^2$

Answer

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$,

$\Rightarrow (pq + 5r)^2\\[1em] \Rightarrow (pq)^2 + 2 \times pq \times 5r + (5r)^2\\[1em] \Rightarrow p^2q^2 + 10pqr + 25r^2$

Hence, $\bm{(pq + 5r)^2 = p^2q^2 + 10pqr + 25r^2}$.

Find the following square, using the identities:

$\left(\dfrac{5}{2}a - \dfrac{3}{5}b\right)^2$

Answer

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

$\Rightarrow \Big(\dfrac{5}{2}a - \dfrac{3}{5}b\Big)^2\\[1em] \Rightarrow \Big(\dfrac{5}{2}a\Big)^2 - 2 \times \dfrac{5}{2}a \times \dfrac{3}{5}b + \Big(\dfrac{3}{5}b\Big)^2\\[1em] \Rightarrow \dfrac{25}{4}a^2 - 3ab + \dfrac{9}{25}b^2$

Hence, $\bm{\Big(\dfrac{5}{2}a - \dfrac{3}{5}b\Big)^2 = \dfrac{25}{4}a^2 - 3ab + \dfrac{9}{25}b^2}$.

Find the following square, using the identities:

$(\sqrt{2}a + \sqrt{3}b)^2$

Answer

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$,

$\Rightarrow (\sqrt{2}a + \sqrt{3}b)^2\\[1em] \Rightarrow (\sqrt{2}a)^2 + 2 \times \sqrt{2}a \times \sqrt{3}b + (\sqrt{3}b)^2\\[1em] \Rightarrow 2a^2 + 2\sqrt{6}ab + 3b^2$

Hence, $\bm{(\sqrt{2}a + \sqrt{3}b)^2 = 2a^2 + 2\sqrt{6}ab + 3b^2}$.

Find the following square, using the identities:

$\left(\dfrac{2x}{3y} - \dfrac{3y}{2x}\right)^2$

Answer

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

$\Rightarrow \Big(\dfrac{2x}{3y} - \dfrac{3y}{2x}\Big)^2\\[1em] \Rightarrow \Big(\dfrac{2x}{3y}\Big)^2 - 2 \times \dfrac{2x}{3y} \times \dfrac{3y}{2x} + \Big(\dfrac{3y}{2x}\Big)^2\\[1em] \Rightarrow \dfrac{4x^2}{9y^2} - 2 + \dfrac{9y^2}{4x^2}$

Hence, $\bm{\Big(\dfrac{2x}{3y} - \dfrac{3y}{2x}\Big)^2 = \dfrac{4x^2}{9y^2} - 2 + \dfrac{9y^2}{4x^2}}$.

Using the identity, $(x + a)(x + b) = x^2 + (a + b)x + ab$, find the following product:

$(x + 7) (x + 3)$

Answer

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$\Rightarrow (x + 7)(x + 3)\\[1em] \Rightarrow x^2 + (7 + 3)x + 7 \times 3\\[1em] \Rightarrow x^2 + 10x + 21$

Hence, $\bm{(x + 7)(x + 3) = x^2 + 10x + 21}$.

Using the identity, $(x + a)(x + b) = x^2 + (a + b)x + ab$, find the following product:

$(3x + 4) (3x - 5)$

Answer

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$\Rightarrow (3x + 4)(3x - 5)\\[1em] \Rightarrow (3x + 4)(3x + (-5))\\[1em] \Rightarrow (3x)^2 + (4 + (-5))(3x) + 4 \times (-5)\\[1em] \Rightarrow 9x^2 - 3x - 20$

Hence, $\bm{(3x + 4)(3x - 5) = 9x^2 - 3x - 20}$.

Using the identity, $(x + a)(x + b) = x^2 + (a + b)x + ab$, find the following product:

$(p^2 + 2q) (p^2 - 3q)$

Answer

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$\Rightarrow (p^2 + 2q)(p^2 - 3q)\\[1em] \Rightarrow (p^2 + 2q)(p^2 + (-3q))\\[1em] \Rightarrow (p^2)^2 + (2q + (-3q))p^2 + 2q \times (-3q)\\[1em] \Rightarrow p^4 - p^2q - 6q^2$

Hence, $\bm{(p^2 + 2q)(p^2 - 3q) = p^4 - p^2q - 6q^2}$.

Using the identity, $(x + a)(x + b) = x^2 + (a + b)x + ab$, find the following product:

$(abc + 3) (abc - 5)$

Answer

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$\Rightarrow (abc + 3)(abc - 5)\\[1em] \Rightarrow (abc + 3)(abc + (-5))\\[1em] \Rightarrow (abc)^2 + (3 + (-5))abc + 3 \times (-5)\\[1em] \Rightarrow a^2b^2c^2 - 2abc - 15$

Hence, $\bm{(abc + 3)(abc - 5) = a^2b^2c^2 - 2abc - 15}$.

Using the identity, $(x + a)(x + b) = x^2 + (a + b)x + ab$, evaluate the following:

203 × 204

Answer

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$\Rightarrow 203 \times 204\\[1em] \Rightarrow (200 + 3)(200 + 4)\\[1em] \Rightarrow (200)^2 + (3 + 4) \times 200 + 3 \times 4\\[1em] \Rightarrow 40000 + 1400 + 12\\[1em] \Rightarrow 41412$

Hence, 203 × 204 = 41412.

Using the identity, $(x + a)(x + b) = x^2 + (a + b)x + ab$, evaluate the following:

8.2 × 8.7

Answer

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$\Rightarrow 8.2 \times 8.7\\[1em] \Rightarrow (8 + 0.2)(8 + 0.7)\\[1em] \Rightarrow (8)^2 + (0.2 + 0.7) \times 8 + 0.2 \times 0.7\\[1em] \Rightarrow 64 + 7.2 + 0.14\\[1em] \Rightarrow 71.34$

Hence, 8.2 × 8.7 = 71.34.

Using the identity, $(x + a)(x + b) = x^2 + (a + b)x + ab$, evaluate the following:

107 × 93

Answer

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$\Rightarrow 107 \times 93\\[1em] \Rightarrow (100 + 7)(100 + (-7))\\[1em] \Rightarrow (100)^2 + (7 + (-7)) \times 100 + 7 \times (-7)\\[1em] \Rightarrow 10000 + 0 - 49\\[1em] \Rightarrow 9951$

Hence, 107 × 93 = 9951.

Using the identity $a^2 - b^2 = (a + b)(a - b)$, find

53² - 47²

Answer

⇒ 53² - 47²

Using the identity $a^2 - b^2 = (a + b)(a - b)$

⇒ (53 + 47)(53 - 47)

⇒ 100 × 6

⇒ 600

Hence, 53² - 47² = 600.

Using the identity $a^2 - b^2 = (a + b)(a - b)$, find

(2.05)² - (0.95)²

Answer

⇒ (2.05)² - (0.95)²

Using the identity $a^2 - b^2 = (a + b)(a - b)$

⇒ (2.05 + 0.95)(2.05 - 0.95)

⇒ 3 × 1.1

⇒ 3.3

Hence, (2.05)² - (0.95)² = 3.3.

Using the identity $a^2 - b^2 = (a + b)(a - b)$, find

(14.3)² - (5.7)²

Answer

⇒ (14.3)² - (5.7)²

Using the identity $a^2 - b^2 = (a + b)(a - b)$

⇒ (14.3 + 5.7)(14.3 - 5.7)

⇒ 20 × 8.6

⇒ 172

Hence, (14.3)² - (5.7)² = 172.

Simplify the following:

$(2x + 5y)^2 + (2x - 5y)^2$

Answer

Using the identities $(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$,

$\Rightarrow (2x + 5y)^2 + (2x - 5y)^2\\[1em] \Rightarrow [(2x)^2 + 2 \times 2x \times 5y + (5y)^2] + [(2x)^2 - 2 \times 2x \times 5y + (5y)^2]\\[1em] \Rightarrow 4x^2 + 20xy + 25y^2 + 4x^2 - 20xy + 25y^2\\[1em] \Rightarrow (4x^2 + 4x^2) + (20xy - 20xy) + (25y^2 + 25y^2)\\[1em] \Rightarrow 8x^2 + 50y^2$

Hence, $\bm{(2x + 5y)^2 + (2x - 5y)^2 = 8x^2 + 50y^2}$.

Simplify the following:

$\left(\dfrac{7}{2}a - \dfrac{5}{2}b\right)^2 - \left(\dfrac{5}{2}a - \dfrac{7}{2}b\right)^2$

Answer

Using the identity $a^2 - b^2 = (a + b)(a - b)$,

$\Rightarrow \Big(\dfrac{7}{2}a - \dfrac{5}{2}b\Big)^2 - \Big(\dfrac{5}{2}a - \dfrac{7}{2}b\Big)^2\\[1em] \Rightarrow \Big[\Big(\dfrac{7}{2}a - \dfrac{5}{2}b\Big) + \Big(\dfrac{5}{2}a - \dfrac{7}{2}b\Big)\Big]\Big[\Big(\dfrac{7}{2}a - \dfrac{5}{2}b\Big) - \Big(\dfrac{5}{2}a - \dfrac{7}{2}b\Big)\Big]\\[1em] \Rightarrow \Big[\dfrac{7}{2}a + \dfrac{5}{2}a - \dfrac{5}{2}b - \dfrac{7}{2}b\Big]\Big[\dfrac{7}{2}a - \dfrac{5}{2}a - \dfrac{5}{2}b + \dfrac{7}{2}b\Big]\\[1em] \Rightarrow [6a - 6b][a + b]\\[1em] \Rightarrow 6(a - b)(a + b)\\[1em] \Rightarrow 6(a^2 - b^2)\\[1em] \Rightarrow 6a^2 - 6b^2$

Hence, $\bm{\Big(\dfrac{7}{2}a - \dfrac{5}{2}b\Big)^2 - \Big(\dfrac{5}{2}a - \dfrac{7}{2}b\Big)^2 = 6a^2 - 6b^2}$.

Simplify the following:

$(p^2 - q^2r)^2 + 2p^2q^2r$

Answer

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

$\Rightarrow (p^2 - q^2r)^2 + 2p^2q^2r\\[1em] \Rightarrow [(p^2)^2 - 2 \times p^2 \times q^2r + (q^2r)^2] + 2p^2q^2r\\[1em] \Rightarrow p^4 - 2p^2q^2r + q^4r^2 + 2p^2q^2r\\[1em] \Rightarrow p^4 + q^4r^2$

Hence, $\bm{(p^2 - q^2r)^2 + 2p^2q^2r = p^4 + q^4r^2}$.

Show that:

$(4x + 7y)^2 - (4x - 7y)^2 = 112xy$

Answer

Taking L.H.S. and using the identity $a^2 - b^2 = (a + b)(a - b)$,

$\Rightarrow (4x + 7y)^2 - (4x - 7y)^2\\[1em] \Rightarrow [(4x + 7y) + (4x - 7y)][(4x + 7y) - (4x - 7y)]\\[1em] \Rightarrow [4x + 7y + 4x - 7y][4x + 7y - 4x + 7y]\\[1em] \Rightarrow [8x][14y]\\[1em] \Rightarrow 112xy\\[1em] \Rightarrow \text{R.H.S.}$

Hence, proved that $\bm{(4x + 7y)^2 - (4x - 7y)^2 = 112xy}$.

Show that:

$\left(\dfrac{3}{7}p - \dfrac{7}{6}q\right)^2 + pq = \dfrac{9}{49}p^2 + \dfrac{49}{36}q^2$

Answer

Taking L.H.S. and using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

$\Rightarrow \Big(\dfrac{3}{7}p - \dfrac{7}{6}q\Big)^2 + pq\\[1em] \Rightarrow \Big(\dfrac{3}{7}p\Big)^2 - 2 \times \dfrac{3}{7}p \times \dfrac{7}{6}q + \Big(\dfrac{7}{6}q\Big)^2 + pq\\[1em] \Rightarrow \dfrac{9}{49}p^2 - \dfrac{2 \times 3 \times 7}{7 \times 6}pq + \dfrac{49}{36}q^2 + pq\\[1em] \Rightarrow \dfrac{9}{49}p^2 - pq + \dfrac{49}{36}q^2 + pq\\[1em] \Rightarrow \dfrac{9}{49}p^2 + \dfrac{49}{36}q^2\\[1em] \Rightarrow \text{R.H.S.}$

Hence, proved that $\bm{\Big(\dfrac{3}{7}p - \dfrac{7}{6}q\Big)^2 + pq = \dfrac{9}{49}p^2 + \dfrac{49}{36}q^2}$.

Show that:

$(p - q)(p + q) + (q - r)(q + r) + (r - p)( r + p) = 0$

Answer

Taking L.H.S. and using the identity $(a + b)(a - b) = a^2 - b^2$,

$\Rightarrow (p - q)(p + q) + (q - r)(q + r) + (r - p)(r + p)\\[1em] \Rightarrow (p^2 - q^2) + (q^2 - r^2) + (r^2 - p^2)\\[1em] \Rightarrow p^2 - q^2 + q^2 - r^2 + r^2 - p^2\\[1em] \Rightarrow (p^2 - p^2) + (-q^2 + q^2) + (-r^2 + r^2)\\[1em] \Rightarrow 0\\[1em] \Rightarrow \text{R.H.S.}$

Hence, proved that $\bm{(p - q)(p + q) + (q - r)(q + r) + (r - p)( r + p) = 0}$.

If $x - \dfrac{1}{x} = 7$, evaluate:

(i) $x^2 + \dfrac{1}{x^2}$

(ii) $x^4 + \dfrac{1}{x^4}$

Answer

(i) Given, $x - \dfrac{1}{x} = 7$

Squaring both sides,

$\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 7^2\\[1em] \Rightarrow x^2 - 2 \times x \times \dfrac{1}{x} + \dfrac{1}{x^2} = 49\\[1em] \Rightarrow x^2 - 2 + \dfrac{1}{x^2} = 49\\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 49 + 2\\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 51$

Hence, $\bm{x^2 + \dfrac{1}{x^2} = 51}$.

(ii) From part (i),

$x^2 + \dfrac{1}{x^2} = 51$

Squaring both sides,

$\Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = 51^2\\[1em] \Rightarrow (x^2)^2 + 2 \times x^2 \times \dfrac{1}{x^2} + \Big(\dfrac{1}{x^2}\Big)^2 = 2601\\[1em] \Rightarrow x^4 + 2 + \dfrac{1}{x^4} = 2601\\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 2601 - 2\\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 2599$

Hence, $\bm{x^4 + \dfrac{1}{x^4} = 2599}$.

If $x^2 + \dfrac{1}{x^2} = 23$, evaluate:

(i) $x + \dfrac{1}{x}$

(ii) $x - \dfrac{1}{x}$

Answer

(i) Given, $x^2 + \dfrac{1}{x^2} = 23$

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + 2 \times x \times \dfrac{1}{x} + \dfrac{1}{x^2}\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 23 + 2\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 25\\[1em] \Rightarrow x + \dfrac{1}{x} = \pm\sqrt{25}\\[1em] \Rightarrow x + \dfrac{1}{x} = \pm 5$

Hence, $\bm{x + \dfrac{1}{x} = \pm5}$.

(ii) Using the identity $(a - b)^2 = a^2 - 2ab + b^2$,

$\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 - 2 \times x \times \dfrac{1}{x} + \dfrac{1}{x^2}\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 23 - 2\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 21\\[1em] \Rightarrow x - \dfrac{1}{x} = \pm\sqrt{21}$

Hence, $\bm{x - \dfrac{1}{x} = \pm \sqrt{21}}$.

If $a + b = 9$ and $ab = 10$, find the value of $a^2 + b^2$

Answer

Given, $a + b = 9$ and $ab = 10$

We know that $(a + b)^2 = a^2 + 2ab + b^2 = a^2 + b^2 + 2ab$

Substituting the values,

$\Rightarrow (9)^2 = a^2 + b^2 + 2 \times 10\\[1em] \Rightarrow 81 = a^2 + b^2 + 20\\[1em] \Rightarrow a^2 + b^2 = 81 - 20\\[1em] \Rightarrow a^2 + b^2 = 61$

Hence, $\bm{a^2 + b^2 = 61}$.

If $a - b = 6$ and $a^2 + b^2 = 42$, find the value of $ab$

Answer

Given, $a - b = 6$ and $a^2 + b^2 = 42$

We know that $(a - b)^2 = a^2 - 2ab + b^2 = a^2 + b^2 - 2ab$

Substituting the values,

$\Rightarrow (6)^2 = 42 - 2ab\\[1em] \Rightarrow 36 = 42 - 2ab\\[1em] \Rightarrow 2ab = 42 - 36\\[1em] \Rightarrow 2ab = 6\\[1em] \Rightarrow ab = 3$

Hence, $\bm{ab = 3}$.

If $a^2 + b^2 = 41$ and $ab = 4$, find the values of

(i) $a + b$

(ii) $a - b$

Answer

Given, $a^2 + b^2 = 41$ and $ab = 4$

(i) We know that $(a + b)^2 = a^2 + b^2 + 2ab$

Substituting the values,

$\Rightarrow (a + b)^2 = 41 + 2 \times 4\\[1em] \Rightarrow (a + b)^2 = 41 + 8\\[1em] \Rightarrow (a + b)^2 = 49\\[1em] \Rightarrow a + b = \pm\sqrt{49}\\[1em] \Rightarrow a + b = \pm 7$

Hence, $\bm{a + b = \pm 7}$.

(ii) We know that $(a - b)^2 = a^2 + b^2 - 2ab$

Substituting the values,

$\Rightarrow (a - b)^2 = 41 - 2 \times 4\\[1em] \Rightarrow (a - b)^2 = 41 - 8\\[1em] \Rightarrow (a - b)^2 = 33\\[1em] \Rightarrow a - b = \pm\sqrt{33}$

Hence, $\bm{a - b = \pm\sqrt{33}}$.