Quadratic Equations — Exercise 5(C)

Discuss the nature of the roots of the following equation without actually solving it:

x² - 8x + 7 = 0

Answer

Comparing x² - 8x + 7 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -8 and c = 7.

We know that,

Discriminant (D) = b² - 4ac = (-8)² - 4 × (1) × (7)

= 64 - 28 = 36; which is positive, and a perfect square.

Hence, the roots are rational and unequal.

Discuss the nature of the roots of the following equation without actually solving it:

6x² + 7x - 10 = 0

Answer

Comparing 6x² + 7x - 10 = 0 with ax² + bx + c = 0 we get,

a = 6, b = 7 and c = -10.

We know that,

Discriminant (D) = b² - 4ac = (7)² - 4 × (6) × (-10)

= 49 -(-240)

= 49 + 240 = 289; which is positive, and a perfect square.

Hence, the roots are rational and unequal.

Discuss the nature of the roots of the following equation without actually solving it:

25x² + 30x + 7 = 0

Answer

Comparing 25x² + 30x + 7 = 0 with ax² + bx + c = 0 we get,

a = 25, b = 30 and c = 7.

We know that,

Discriminant (D) = b² - 4ac = (30)² - 4 × 25 × 7

= 900 - 700 = 200; which is positive, but not a perfect square.

Hence, the roots are irrational and unequal.

Discuss the nature of the roots of the following equation without actually solving it:

15x² - 28 = x

Answer

⇒ 15x² - 28 = x

⇒ 15x² - x - 28 = 0

Comparing 15x² - x - 28 = 0 with ax² + bx + c = 0 we get,

a = 15, b = -1 and c = -28.

We know that,

Discriminant (D) = b² - 4ac = (1)² - 4 × 15 × -28

= 1 - (-1680) = 1 + 1680 = 1681; which is positive and is a perfect square.

Hence, the roots are rational and unequal.

Discuss the nature of the roots of the following equation without actually solving it:

16x² = 24x + 1

Answer

⇒ 16x² = 24x + 1

⇒ 16x² - 24x - 1 = 0

Comparing 16x² - 24x - 1 = 0 with ax² + bx + c = 0 we get,

a = 16, b = -24 and c = -1.

We know that,

Discriminant (D) = b² - 4ac

= (-24)² - 4 × 16 × -1

= 576 - (-64) = 576 + 64 = 640; which is positive and is not a perfect square.

Hence, the roots are irrational and unequal.

Discuss the nature of the roots of the following equation without actually solving it:

2x² - $2\sqrt{6}x$ + 3 = 0

Answer

Comparing 2x² - $2\sqrt{6}x$ + 3 = 0 with ax² + bx + c = 0 we get,

a = 2, b = $-2\sqrt{6}$ and c = 3.

We know that,

Discriminant (D) = b² - 4ac = $(-2\sqrt{6})^2$ - 4 × 2 × 3

= (4 × 6) - 24 = 24 - 24 = 0;

Since, b is irrational and discriminant equals to zero.

Hence, the roots are irrational and equal.

Discuss the nature of the roots of the following equation without actually solving it:

2x² + 2x + 3 = 0

Answer

Comparing 2x² + 2x + 3 = 0 with ax² + bx + c = 0 we get,

a = 2, b = 2 and c = 3.

We know that,

Discriminant (D) = b² - 4ac = (2)² - 4 × 2 × 3

= 4 - 24 = -20; which is negative.

Hence, the roots are imaginary and unequal.

Discuss the nature of the roots of the following equation without actually solving it:

2x² - 5x - 4 = 0

Answer

Comparing 2x² - 5x - 4 = 0 with ax² + bx + c = 0 we get,

a = 2, b = -5 and c = -4.

We know that,

Discriminant (D) = b² - 4ac = (-5)² - 4 × 2 × -4

= 25 - (-32) = 25 + 32 = 57 ; which is positive and is not a perfect square.

Hence, the roots are irrational and unequal.

Discuss the nature of the roots of the following equation without actually solving it:

5x² - 13x - 6 = 0

Answer

Comparing 5x² - 13x - 6 = 0 with ax² + bx + c = 0 we get,

a = 5, b = -13 and c = -6.

We know that,

Discriminant (D) = b² - 4ac = (-13)² - 4 × 5 × -6

= 169 - (-120) = 169 + 120 = 289 ; which is positive and is a perfect square.

Hence, the roots are rational and unequal.

Discuss the nature of the roots of the following equation without actually solving it:

9x² - 6x + 1 = 0

Answer

Comparing 9x² - 6x + 1 = 0 with ax² + bx + c = 0 we get,

a = 9, b = -6 and c = 1.

We know that,

Discriminant (D) = b² - 4ac = (-6)² - 4 × 9 × 1

= 36 - 36 = 0.

Hence, the roots are rational and equal.

Discuss the nature of the roots of the following equation without actually solving it:

3x² - 2x + 5 = 0

Answer

Comparing 3x² - 2x + 5 = 0 with ax² + bx + c = 0 we get,

a = 3, b = -2 and c = 5.

We know that,

Discriminant (D)= b² - 4ac = (-2)² - 4 × 3 × 5

= 4 - 60 = -56; which is negative.

Hence, the roots are imaginary and unequal.

Discuss the nature of the roots of the following equation without actually solving it:

x² + $2\sqrt{3}x$ - 1 = 0

Answer

Comparing x² + $2\sqrt{3}x$ - 1 = 0 with ax² + bx + c = 0 we get,

a = 1, b = $2\sqrt{3}$ and c = -1.

We know that,

Discriminant = D = b² - 4ac = $(2\sqrt{3})^2$ - 4 × 1 × -1

= (2 × 3) + 4 = 10 ; which is positive and is not a perfect square.

Since, b is irrational and discriminant is greater than zero.

Hence, the roots are irrational and unequal.

Find the values of k for which the following equation has equal roots:

9x² + kx + 1 = 0

Answer

Comparing 9x² + kx + 1 = 0 with ax² + bx + c = 0 we get,

a = 9, b = k and c = 1.

Since equations has equal roots,

∴ D = 0

⇒ (k)² - 4 × 9 × 1 = 0

⇒ k² - 36 = 0

⇒ k² = 36

⇒ k = $\sqrt{36}$

⇒ k = ± 6

Hence, k = {6, -6}.

Find the values of k for which the following equation has equal roots:

x² - 2kx + 7k - 12 = 0

Answer

Comparing x² - 2kx + 7k - 12 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -2k and c = (7k - 12).

Since equations has equal roots,

∴ D = 0

⇒ (-2k)² - 4 × 1 × (7k - 12) = 0

⇒ 4k² - (28k - 48) = 0

⇒ 4k² - 28k + 48 = 0

⇒ 4k² - 16k - 12k + 48 = 0

⇒ 4k(k - 4) - 12(k - 4) = 0

⇒ (k - 4)(4k - 12) = 0

⇒ (k - 4) = 0 or (4k - 12) = 0      [Using Zero-product rule]

⇒ k = 4 or 4k = 12

⇒ k = 4 or k = $\dfrac{12}{4}$

⇒ k = 4 or k = 3.

Hence, k = {4, 3}.

Find the values of k for which the following equation has equal roots:

(3k + 1)x² + 2(k + 1)x + k = 0

Answer

Comparing (3k + 1)x² + 2(k + 1)x + k = 0 with ax² + bx + c = 0 we get,

a = (3k + 1), b = 2(k + 1) and c = k.

Since equations has equal roots,

∴ D = 0

⇒ [2(k + 1)]² - 4 × (3k + 1) × (k) = 0

⇒ 4(k + 1)² - (12k + 4) × (k) = 0

⇒ 4[(k)² + (1)² + 2 × k × 1] - (12k² + 4k) = 0

⇒ 4(k² + 1 + 2k) - 12k² - 4k = 0

⇒ 4k² + 4 + 8k - 12k² - 4k = 0

⇒ -8k² + 4k + 4 = 0

⇒ -8k² + 8k - 4k + 4 = 0

⇒ -8k(k - 1) - 4(k - 1) = 0

⇒ (k - 1)(-8k - 4) = 0

⇒ (k - 1) = 0 or (-8k - 4) = 0      [Using Zero-product rule]

⇒ k = 1 or -8k = 4

⇒ k = 1 or k = $\dfrac{4}{-8}$

⇒ k = 1 or k = $-\dfrac{1}{2}$

Hence, k = $\Big\lbrace-\dfrac{1}{2}, 1\Big\rbrace$.

Find the values of k for which the following equation has equal roots:

x² - 2(5 + 2k)x + 3(7 + 10k) = 0

Answer

Comparing x² - 2(5 + 2k)x + 3(7 + 10k) = 0 with ax² + bx + c = 0 we get,

a = 1, b = -2(5 + 2k) and c = 3(7 + 10k).

Since equations has equal roots,

∴ D = 0

⇒ [-2(5 + 2k)]² - 4 × 1 × 3(7 + 10k) = 0

⇒ 4(5 + 2k)² - 12(7 + 10k) = 0

⇒ 4[(5)² + (2k)² + 2 × 5 × 2k] - (84 + 120k) = 0

⇒ 4(25 + 4k² + 20k) - 84 - 120k = 0

⇒ 100 + 16k² + 80k - 84 - 120k = 0

⇒ 16k² - 40k + 16 = 0

⇒ 16k² - 8k - 32k + 16 = 0

⇒ 8k(2k - 1) - 16(2k - 1) = 0

⇒ (2k - 1)(8k - 16)= 0

⇒ (2k - 1) = 0 or (8k - 16)= 0      [Using Zero-product rule]

⇒ 2k = 1 or 8k = 16

⇒ k = $\dfrac{1}{2}$ or k = $\dfrac{16}{8}$

⇒ k = $\dfrac{1}{2}$ or k = 2

Hence, k = $\Big\lbrace2 , \dfrac{1}{2}\Big\rbrace$.

Find the values of k for which the following equation has equal roots:

(k + 1)x² + 2(k + 3)x + (k + 8) = 0

Answer

Comparing (k + 1)x² + 2(k + 3)x + (k + 8) = 0 with ax² + bx + c = 0 we get,

a = (k + 1), b = 2(k + 3) and c = (k + 8).

Since equations has equal roots,

∴ D = 0

⇒ [2(k + 3)]² - 4 × (k + 1) × (k + 8) = 0

⇒ 4(k + 3)² - (4k + 4) × (k + 8) = 0

⇒ 4[(k)² + (3)² + 2 × k × 3] - (4k² + 32k + 4k + 32) = 0

⇒ 4(k² + 9 + 6k) - (4k² + 36k + 32) = 0

⇒ 4k² + 36 + 24k - 4k² - 36k - 32 = 0

⇒ -12k + 4 = 0

⇒ -12k = -4

⇒ k = $\dfrac{-4}{-12}$

⇒ k = $\dfrac{1}{3}$.

Hence, k = $\Big\lbrace\dfrac{1}{3}\Big\rbrace$.

Find the values of k for which the following equation has equal roots:

kx² + kx + 1 = -4x² - x

Answer

⇒ kx² + kx + 1 = -4x² - x

⇒ kx² + kx + 1 + 4x² + x = 0

⇒ kx² + 4x² + kx + x + 1 = 0

⇒ x² (k + 4) + x(k + 1) + 1 = 0

Comparing x² (k + 4) + x(k + 1) + 1 = 0 with ax² + bx + c = 0 we get,

a = (k + 4), b = (k + 1) and c = 1.

Since equations has equal roots,

∴ D = 0

⇒ (k + 1)² - 4.(k + 4).1 = 0

⇒ [(k)² + (1)² + 2 × k × 1] - (4k + 16) = 0

⇒ k² + 1 + 2k - 4k - 16 = 0

⇒ k² - 2k - 15 = 0

⇒ k² - 5k + 3k - 15 = 0

⇒ k(k - 5) + 3(k - 5) = 0

⇒ (k - 5)(k + 3) = 0

⇒ (k - 5) = 0 or (k + 3) = 0      [Using Zero-product rule]

⇒ k = 5 or k = -3

Hence, k = {5 , -3}.

Find the values of k for which the following equation has equal roots:

3kx² = 4(kx - 1)

Answer

⇒ 3kx² = 4(kx - 1)

⇒ 3kx² = 4kx - 4

⇒ 3kx² - 4kx + 4 = 0

Comparing 3kx² - 4kx + 4 = 0 with ax² + bx + c = 0 we get,

a = 3k, b = -4k and c = 4.

Since equations has equal roots,

∴ D = 0

⇒ (-4k)² - 4 × (3k) × 4 = 0

⇒ 16k² - 48k = 0

⇒ 16k(k - 3) = 0

⇒ 16k = 0 or (k - 3) = 0      [Using Zero-product rule]

⇒ k = 0 or k = 3

Hence, k = {0, 3}.

Find the values of k for which the following equation has equal roots:

x² + 4kx + (k² - k + 2) = 0

Answer

Comparing x² + 4kx + (k² - k + 2) = 0 with ax² + bx + c = 0 we get,

a = 1, b = 4k and c = (k² - k + 2).

Since equations has equal roots,

∴ D = 0

⇒ (4k)² - 4.1.(k² - k + 2) = 0

⇒ 16k² - (4k² - 4k + 8) = 0

⇒ 16k² - 4k² + 4k - 8 = 0

⇒ 12k² + 4k - 8 = 0

⇒ 12k² + 12k - 8k - 8 = 0

⇒ 12k(k + 1) - 8(k + 1) = 0

⇒ (k + 1)(12k - 8) = 0

⇒ (k + 1) = 0 or (12k - 8) = 0      [Using Zero-product rule]

⇒ k = -1 or 12k = 8

⇒ k = -1 or k = $\dfrac{8}{12}$

⇒ k = -1 or k = $\dfrac{2}{3}$.

Hence, k = $\Big\lbrace-1,\dfrac{2}{3}\Big\rbrace$.

Show that the equation x² + ax - 1 = 0 has real and distinct roots for all real values of a.

Answer

Comparing x² + ax - 1 = 0 with ax² + bx + c = 0 we get,

a = 1, b = a and c = -1.

We know that,

Discriminant (D) = b² - 4ac = (a)² - 4.(1).(-1) = a² + 4

Since a² ≥ 0 for all real values of a, we have D = a² + 4 > 0

The equation has real and distinct roots for all real values of a.

Hence, equation x² + ax - 1 = 0 has real and distinct roots for all real values of a.

Show that the equation 3x² + 7x + 8 = 0 is not true for any real value of x.

Answer

Comparing 3x² + 7x + 8 = 0 with ax² + bx + c = 0 we get,

a = 3, b = 7 and c = 8.

We know that,

Discriminant (D) = b² - 4ac = (7)² - 4.(3).(8) = 49 - 96 = - 47; which is negative.

The equation has imaginary and unequal roots for all real values of x.

Hence, the equation 3x² + 7x + 8 = 0 is not true for any real value of x.

If the roots of the equation (c² - ab)x² - 2(a² - bc)x + (b² - ac) = 0 are real and equal, show that either a = 0 or a³ + b³ + c³ = 3abc.

Answer

Comparing (c² - ab)x² - 2(a² - bc)x + (b² - ac) = 0 with ax² + bx + c = 0 we get,

a = (c² - ab), b = -2(a² - bc) and c = (b² - ac).

Since equations has equal roots,

∴ D = 0

⇒ [-2(a² - bc)]² - 4 × (c² - ab) ×(b² - ac) = 0

⇒ 4(a² - bc)² - 4 × (c²b² - ac³ - ab³ + a²bc ) = 0

⇒ 4[(a²)² + (bc)² - 2 × a² × bc] - (4c²b² - 4ac³ - 4ab³ + 4a²bc) = 0

⇒ 4(a⁴ + b²c² - 2a²bc) - 4c²b² + 4ac³ + 4ab³ - 4a²bc = 0

⇒ 4a⁴ + 4b²c² - 8a²bc - 4c²b² + 4ac³ + 4ab³ - 4a²bc = 0

⇒ 4a⁴ + 4b²c² - 4c²b² - 8a²bc - 4a²bc + 4ac³ + 4ab³ = 0

⇒ 4a⁴ - 12a²bc + 4ac³ + 4ab³ = 0

⇒ 4a(a³ - 3abc + c³ + b³) = 0

⇒ 4a = 0 or (a³ + c³ + b³ - 3abc) = 0      [Using Zero-product rule]

⇒ a = 0 or a³ + c³ + b³ = 3abc

Hence, proved a = 0 or a³ + b³ + c³ = 3abc.

If a, b, c ∈ R, show that the roots of the equation (a - b)x² + (b + c - a)x - c = 0 are rational.

Answer

(a - b)x² + (b + c - a)x - c = 0, a ≠ b

Comparing (a - b)x² + (b + c - a)x - c = 0 with ax² + bx + c = 0 we get,

a = (a - b), b = (b + c - a) and c = -c.

We know that,

Discriminant (D) = b² - 4ac

= (b + c - a)² - 4 × (a - b) × (-c)

= [(b + c) - a]² - 4 × (-ac + bc)

= [(b + c)² + (a)² - 2 × (b + c) × (a)] - (-4ac + 4bc)

= [(b)² + (c)² + 2 × b × c + a² - 2 × (ab + ac)] + 4ac - 4bc

= (b² + c² + 2bc + a² - 2ab - 2ac) + 4ac - 4bc

= a² + b² + c² + 2bc - 2ab - 2ac + 4ac - 4bc

= a² + b² + c² + 2bc - 4bc - 2ab - 2ac + 4ac

= a² + b² + c² - 2bc - 2ab + 2ac

= a² + c² + 2ac + b² - 2ab - 2bc

= a² + c² + 2ac + b² - 2.(a + c).b

= (a + c - b)²

Thus D = (a + c - b)², which is a perfect square.

The equation has rational roots. The roots are unequal if b ≠ a + c and equal if b = a + c (as then discriminant equals to zero).

Hence, (a - b)x² + (b + c - a)x - c = 0 has rational roots. The roots are unequal if b ≠ a + c and equal if b = a + c.

If a, b, c are rational, prove that the roots of the equation (b - c)x² + (c - a)x + (a - b) = 0 are also rational.

Answer

Comparing (b - c)x² + (c - a)x + (a - b) = 0 with ax² + bx + c = 0 we get,

a = (b - c), b = (c - a) and c = (a - b).

We know that,

Discriminant (D) = b² - 4ac

= (c - a)² - 4 × (b - c) × (a - b)

= [(c)² + (a)² - 2 × c × a] - 4 × (ba - b² - ac + bc)

= (c² + a² - 2ac) -(4ba - 4b² - 4ac + 4bc)

= c² + a² - 2ac - 4ba + 4b² + 4ac - 4bc

= c² + a² - 2ac + 4ac - 4ba + 4b² - 4bc

= c² + a² + 2ac - 4ba + 4b² - 4bc

= c² + a² + 2ac + (2b)² - 2.(c + a).2b

= (a + c - 2b)²

Thus, D = (a + c - 2b)², which is a perfect square.

The equation has rational roots.

Hence, proved that (b - c)x² + (c - a)x + (a - b) = 0 has rational roots.