Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Theory of Equations Solutions Exercise 4(b) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2A Theory of Equations Solutions Exercise 4(b)

I.

Question 1.

Solve x^{3} – 3x^{2} – 16x + 48 = 0, given that the sum of two roots is zero.

Solution:

Let α, β, γ are the roots of x^{3} – 3x^{2} – 16x + 48 = 0

α + β + γ = 3

Given α + β = 0 (∵ Sum of two roots is zero)

∴ γ = 3

i.e. x – 3 is a factor of x^{3} – 3x^{2} – 16x + 48 = 0

x^{2} – 16 = 0

⇒ x^{2} = 16

⇒ x = ±4

∴ The roots are -4, 4, 3

Question 2.

Find the condition that x^{3} – px^{2} + qx – r = 0 may have the sum of its roots zero.

Solution:

Let α, β, γ be the roots of x^{3} – px^{2} + qx – r = 0

α + β + γ = p ………(1)

αβ + βγ + γα = q ………..(2)

αβγ = r ………..(3)

Given α + β = 0 (∵ Sum of two roots is zero)

From (1), γ = p

∴ γ is a root of x^{3} – px^{2} + qx – r = 0

γ^{3} – pγ^{2} + qγ – r = 0

But γ = p

p^{3} – p(p^{2}) + q(p) – r = 0

p^{3} – p^{2} + qp – r = 0

∴ qp = r is the required condition.

Question 3.

Given that the roots of x^{3} + 3px^{2} + 3qx + r = 0 are in

(i) A.P., show that 2p^{3} – 3qp + r = 0

(ii) G.P., show that p^{3}r = q^{3}

(iii) H.P., show that 2q^{3} = r(3pq – r)

Solution:

Given equation is x^{3} + 3px^{2} + 3qx + r = 0

(i) The roots are in A.P.

Suppose a – d, a, a + d are the roots

Sum = a – d + a + a + d = -3p

⇒ 3a = -3p

⇒ a = -p ……….(1)

∵ ‘a’ is a root of x^{3} + 3px^{2} + 3qx + r = 0

⇒ a^{3} + 3pa^{2} + 3qa + r = 0

But a = -p

⇒ p^{3} + 3p(-p)^{2} + 3q(-p) + r = 0

⇒ 2p^{3} – 3pq + r = 0 is the required condition

(ii) The roots are in G.P.

Suppose the roots be \(\frac{a}{R}\), a, aR

Given (\(\frac{a}{R}\)) (a) (aR) = -r

⇒ a = -r

⇒ a = (-r)^{1/3}

∵ ‘a’ is a root of x^{3} + 3px^{2} + 3qx + r = 0

⇒ (-r^{1/3})^{3} + 3p(-r^{1/3})^{2} + 3q(-r^{1/3}) + r = 0

⇒ -r + 3pr^{2/3} – 3qr^{1/3} + r = 0

pr^{2/3} = qr^{1/3}

⇒ pr^{1/3} = q

⇒ p^{1/3}r = q is the required condition

(iii) The roots of x^{3} + 3px^{2} + 3qx + r = 0 are in H.P.

x^{3} + 3px^{2} + 3qx + r = 0 ……..(1)

Let y = \(\frac{1}{x}\) so that \(\frac{1}{y^{3}}+\frac{3 p}{y^{2}}+\frac{3 q}{y}+r\) = 0

Roots of ry^{3} + 3qy^{2} + 3py + 1 = 0 are in A.P.

ry^{3} + 3qy^{2} + 3py + 1 = 0 ……..(2)

Suppose a – d, a, a + d be the roots of (2)

Sum = a – d + a + a + d = \(-\frac{3 q}{r}\)

3a = \(-\frac{3 q}{r}\)

a = \(-\frac{q}{r}\) ………(1)

∵ ‘a’ is a root of ry^{3} + 3qy^{2} + 3py + 1 = 0

⇒ ra^{3} + 3qa^{2} + 3pa + 1 = 0

But a = \(-\frac{q}{r}\)

⇒ \(r\left(-\frac{q}{r}\right)^{3}+3 q\left(-\frac{q}{r}\right)^{2}+3 p\left(-\frac{q}{r}\right)+1=0\)

⇒ \(\frac{-q^{3}}{r^{2}}+\frac{3 q^{3}}{r^{2}}-\frac{3 p q}{r}+1=0\)

⇒ -q^{3} + 3q^{3} – 3pqr + r^{2} = 0

⇒ 2q^{3} = r(3pq – r) is the required condition.

Question 4.

Find the condition that x^{3} – px^{2} + qx – r = 0 may have the roots in G.P.

Solution:

Let \(\frac{a}{R}\), a, aR be the roots of x^{3} – px^{2} + qx – r = 0

The product of the roots = \(\frac{a}{R}\) . a . aR = a^{3}

product of the roots = r

⇒ a = r^{1/3}

∵ a is a root of x^{3} – px^{2} + qx – r = 0

⇒ a^{3} – pa^{2} + qa – r = 0

But a = r^{1/3}

⇒ (r^{1/3})^{3} – p(r^{1/3})^{2} + q(r^{1/3}) – r = 0

⇒ r – p . r^{2/3} + q . r^{1/3} – r = 0

⇒ p . r^{2/3} = qr^{1/3}

By cubing on both sides

⇒ p^{3}r^{2} = q^{3}r

⇒ p^{3}r = q^{3} is the required condition

II.

Question 1.

Solve 9x^{3} – 15x^{2} + 7x – 1 = 0, given that two of its roots are equal.

Solution:

Suppose α, β, γ are the roots of 9x^{3} – 15x^{3} + 7x – 1 = 0

α + β + γ = \(\frac{15}{9}=\frac{5}{3}\)

αβ + βγ + γα = \(\frac{7}{9}\)

αβγ = \(\frac{1}{9}\)

Given α = β (∵ two of its roots are equal)

2α + γ = \(\frac{5}{3}\)

⇒ γ = \(\frac{5}{3}\) – 2α

α^{2} + 2αγ = \(\frac{7}{9}\)

⇒ α^{2} + 2α (\(\frac{5}{3}\) – 2α) = \(\frac{7}{9}\)

⇒ α^{2} + \(\frac{2 \alpha(5-6 \alpha)}{3}=\frac{7}{9}\)

⇒ 9α^{2} + 6α(5 – 6α) = 7

⇒ 9α^{2} + 30α – 36α^{2} = 7

⇒ 27α^{2} – 30α + 7 = 0

⇒ (3α – 1)(9α – 7) = 0

⇒ α = \(\frac{1}{3}\) or \(\frac{7}{9}\)

The roots are \(\frac{1}{3}\), \(\frac{1}{3}\), 1

Question 2.

Given that one root of 2x^{3} + 3x^{2} – 8x + 3 = 0 is double of another root, find the roots of the equation.

Solution:

Suppose α, β, γ are the roots of 2x^{3} + 3x^{2} – 8x + 3 = 0

α + β + γ = \(-\frac{3}{2}\) ……..(1)

αβ + βγ + γα = -4 ……..(2)

αβγ = \(-\frac{3}{2}\)

Given α = 2β (∵ one root is double the other)

Substituting in (1)

3β + γ = \(-\frac{3}{2}\)

⇒ γ = \(-\frac{3}{2}\) – 3β …….(4)

Substituting in (2)

αβ + γ(α + β) = -4

⇒ 2β^{2} + 3βγ = -4

⇒ 2β^{2} + 3β(\(-\frac{3}{2}\) – 3β) = -4

⇒ 2β^{2} – \(\frac{3 \beta(3+6 \beta)}{2}\) = -4

⇒ 4β^{2} – 9β – 18β^{2} = -8

⇒ 14β^{2} + 9β – 8 = 0

⇒ (2β – 1)(7β + 8) = 0

⇒ 2β – 1 = 0 or 7β + 8 = 0

⇒ β = \(\frac{1}{2}\) or β = \(-\frac{8}{7}\)

∴ The roots are \(\frac{1}{2}\), 1 and -3

Question 3.

Solve x^{3} – 9x^{2} + 14x + 24 = 0, given that two of the roots are in the ratio 3 : 2.

Solution:

Suppose α, β, γ are the roots of x^{3} – 9x^{2} + 14x + 24 = 0

α + β + γ = 9 ………(1)

αβ + βγ + γα = 14 ……….(2)

αβγ = -24 ……….(3)

∵ two roots are in the ratio 3 : 2

Let α : β = 3 : 2

⇒ β = \(\frac{2 \alpha}{3}\)

Substituting in (1)

\(\frac{5 \alpha}{3}\) + γ = 9

⇒ γ = 9 – \(\frac{5 \alpha}{3}\) ………(4)

Substituting in (2)

⇒ \(\frac{2}{3}\) α^{2} + γ(α + β) = 14

⇒ \(\frac{2}{3} \alpha^{2}+\left(9-\frac{5 \alpha}{3}\right) \cdot \frac{5 \alpha}{3}\) = 14

⇒ 2α^{2} + 5α(9 – \(\frac{5 \alpha}{3}\)) = 42

⇒ 2α^{2} + 5α \(\frac{(27-5 \alpha)}{3}\) = 42

⇒ 6α^{2} + 135α – 25α^{2} = 126

⇒ 19α^{2} – 135α + 126 = 0

⇒ 19α^{2} – 114α – 21α + 126 = 0

⇒ 19α(α – 6) – 21(α – 6) = 0

⇒ (19α – 21)(α – 6) = 0

⇒ 19α – 21 = 0 or α – 6 = 0

⇒ α = \(\frac{21}{19}\) or α = 6

Case (i): α = 6

β = \(\frac{2 \alpha}{3}\)

= \(\frac{2}{3}\) × 6

= 4

γ = 9 – \(\frac{5 \alpha}{3}\)

= 9 – \(\frac{5}{3}\) × 6

= 9 – 10

= -1

α = 6, β = 4, γ = -1 satisfy αβγ = -24

∴ The roots are 6, 4, -1

Case (ii): α = \(\frac{21}{19}\)

β = \(\frac{2}{3} \times \frac{21}{19}=\frac{14}{19}\)

γ = 9 – \(\frac{5 \alpha}{3}\)

= 9 – \(\frac{5}{3} \cdot \frac{21}{19}\)

= \(\frac{136}{19}\)

These values do not satisfy αβγ= -24

∴ The roots are 6, 4, -1.

Question 4.

Solve the following equations, given that the roots of each are in A.P.

(i) 8x^{3} – 36x^{2} – 18x + 81 = 0

Solution:

Given the roots of 8x^{3} – 36x^{2} – 18x + 81 = 0 are in A.P.

Let the roots be a – d, a, a + d

Sum of the roots = \(\frac{36}{8}\)

⇒ a – d + a + a + d = \(\frac{9}{2}\)

⇒ 3a = \(\frac{9}{2}\)

⇒ a = \(\frac{3}{2}\)

∴ (x – \(\frac{3}{2}\)) is a factor of 8x^{3} – 36x^{2} – 18x + 81 = 0

⇒ 8x^{2} – 24x – 54 = 0

⇒ 4x^{2} – 12x – 27 = 0

⇒ 4x^{2} – 18x + 6x – 27 = 0

⇒ 2x(2x – 9) + 3(2x – 9) = 0

⇒ (2x + 3) (2x – 9) = 0

⇒ x = \(-\frac{3}{2}, \frac{9}{2}\)

∴ The roots are \(-\frac{3}{2}, \frac{3}{2}, \frac{9}{2}\)

(ii) x^{3} – 3x^{2} – 6x + 8 = 0

Solution:

The roots of x^{3} – 3x^{2} – 6x + 8 = 0 are in A.P

Suppose a – d, a, a + d be the roots

Sum = a – d + a + a + d = 3

⇒ 3a = 3

⇒ a = 1

∴ (x – 1) is a factor of x^{3} – 3x^{2} – 6x + 8 = 0

⇒ x^{2} – 2x – 8 = 0

⇒ x^{2} – 4x + 2x – 8 = 0

⇒ x(x – 4)+ 2(x – 4) = 0

⇒ (x – 4) (x + 2) = 0

⇒ x = 4, -2

∴ The roots are -2, 1, 4

Question 5.

Solve the following equations, given that the roots of each are in G.P.

(i) 3x^{3} – 26x^{2} + 52x – 24 = 0

Solution:

Given equation is 3x^{3} – 26x^{2} + 52x – 24 = 0

The roots are in G.P.

Suppose \(\frac{a}{r}\), a, ar are the roots.

Product = \(\frac{a}{r}\) . a . ar = \(-\left(-\frac{24}{3}\right)\)

⇒ a^{3} = 8

⇒ a = 2

∴ (x – 2) is a factor of 3x^{3} – 26x^{2} + 52x – 24

Hint: 3 × 12 = 3 ×6 × 2 = (-18)(-2)

⇒ 3x^{2} – 20x + 12 = 0

⇒ 3x^{2} – 18x – 2x + 12 = 0

⇒ 3x(x – 6) – 2(x – 6) = 0

⇒ (3x – 2) (x – 6) = 0

⇒ x = \(\frac{2}{3}\), 6

∴ The roots are \(\frac{2}{3}\), 2, 6.

(ii) 54x^{3} – 39x^{2} – 26x + 16 = 0

Solution:

Given equation is 54x^{3} – 39x^{2} – 26x + 16 = 0

The roots are in G.P.

Suppose \(\frac{a}{r}\), a, ar be the roots.

Product = \(\frac{a}{r}\) . a . ar = \(-\frac{16}{54}\)

Hint: 18 × 8 = 9 × 2 × 8 = (-9) (-16)

⇒ 54x^{2} – 75x + 24 = 0

⇒ 18x^{2} – 25x + 8 = 0

⇒ 18x^{2} – 9x – 16x + 8 = 0

⇒ 9x(2x – 1) – 8(2x- 1) = 0

⇒ (9x – 8) (2x – 1) = 0

⇒ x = \(\frac{8}{9}, \frac{1}{2}\)

∴ The roots are \(\frac{8}{9},-\frac{2}{3}, \frac{1}{2}\)

Question 6.

Solve the following equations, given that the roots of each are in H.P.

(i) 6x^{3} – 11x^{2} + 6x – 1 = 0

Solution:

Given equation is 6x^{3} – 11x^{2} + 6x – 1 = 0 …….(1)

Put y = \(\frac{1}{x}\) so that \(\frac{6}{\mathrm{y}^{3}}-\frac{11}{\mathrm{y}^{2}}+\frac{6}{\mathrm{y}}-1\) = 0

⇒ 6 – 11y + 6y^{2} – y^{3} = 0

⇒ y^{3} – 6y^{2} + 11y – 6 = 0 ………(2)

Roots of (1) are in H.P.

⇒ Roots of (2) are in A.P.

Let a – d, a, a + d be the roots of (2)

Sum = a – d + a + a + d = 6

⇒ 3a = 6

⇒ a = 2

Product = a(a^{2} – d^{2}) = 6

⇒ 2(4 – d^{2}) = 6

⇒ 4 – d^{2} = 3

⇒ d^{2} = 1

⇒ d = 1

a – d = 2 – 1 = 1,

a = 2

a + d = 2 + 1 = 3

The roots of (2) are 1, 2, 3

The roots of (1) are 1, \(\frac{1}{2}\), \(\frac{1}{3}\)

(ii) 15x^{3} – 23x^{2} + 9x – 1 = 0

Solution:

Given equation is 15x^{3} – 23x^{2} + 9x – 1 = 0 …….(1)

Put y = \(\frac{1}{x}\) so that \(\frac{15}{y^{3}}-\frac{23}{y^{2}}+\frac{9}{y}-1\) = 0

⇒ 15 – 23y + 9y^{2} – y^{3} = 0

⇒ y^{3} – 9y^{2} + 23y – 15 = 0 ………(2)

Roots of (1) are in H.P. So that roots of (2) are in A.P.

Let a – d, a, a + d be the roots of (2)

Sum = a – d + a + a + d = 9

⇒ 3a = 9

⇒ a = 3

Product = a(a^{2} – d^{2}) = 15

⇒ 3(9 – d^{2}) = 15

⇒ 9 – d^{2} = 5

⇒ d^{2} = 4

⇒ d = 2

a – d = 3 – 2 = 1

a = 3

a + d = 3 + 2 = 5

Roots of (2) are 1, 3, 5

Hence roots of (1) are 1, \(\frac{1}{3}\), \(\frac{1}{5}\)

Question 7.

Solve the following equations, given that they have multiple roots.

(i) x^{4} – 6x^{3} + 13x^{2} – 24x + 36 = 0

Solution:

(i) Let f(x) = x^{4} – 6x^{3} + 13x^{2} – 24x + 36

⇒ f'(x) = 4x^{3} – 18x^{2} + 26x – 24

⇒ f'(3) = 4(27) – 18(9) + 26(3) – 24

⇒ f'(3) = 108 – 162 + 78 – 24

⇒ f'(3) = 0

f(3) = 81 – 162 + 117 – 72 + 36 = 0

Hint: Choose the value of x from the factors of the G.C.D of constant terms in f(x) and f'(x).

∴ x – 3 is a factor of f'(x) and f(x)

∴ 3 is the repeated foot of f(x)

x^{2} + 4 = 0

⇒ x = ±2i

∴ The roots of the given equation are 3, 3, ±2i

(ii) 3x^{4} + 16x^{3} + 24x^{2} – 16 = 0

Solution:

Let f(x) = 3x^{4} + 16x^{3} + 24x^{2} – 16

f(x) = 12x^{3} + 48x^{2} + 48x

= 12x(x^{2} + 4x + 4)

= 12x (x + 2)^{2}

f'(-2) = 0

f(-2) = 3(16) + 16(-8) + 24(4) – 16 = 0

∴ x + 2 is a factor of f'(x) and f(x)

∴ -2 is a multiple root of f(x) = 0

3x^{2} + 4x – 4 = 0

⇒ 3x^{2} + 6x – 2x – 4 = 0

⇒ 3x(x + 2) – 2(x + 2) = 0

⇒ (x + 2) (3x – 2) = 0

⇒ x = -2, \(\frac{2}{3}\)

∴ The roots of the given equation are -2, -2, -2, \(\frac{2}{3}\)

III.

Question 1.

Solve x^{4} + x^{3} – 16x^{2} – 4x + 48 = 0, given that the product of two of the roots is 6.

Solution:

Suppose α, β, γ, δ are the roots of

x^{4} + x^{3} – 16x^{2} – 4x + 48 = 0 ………..(1)

Sum of the roots = -1

⇒ α + β + γ + δ = -1

and Product of the roots = αβγδ = 48

∵ Product of two roots is 6

Let αβ = 6

From (1), γδ = \(\frac{48}{\alpha \beta}=\frac{48}{6}\) = 8

Let α + β = p and γ + δ = q

The equation having roots α, β is x^{2} – (α + β) x + αβ = 0

⇒ x^{2} – px + 6 = 0 ………..(2)

The equation having the roots γ, δ is x^{2} – (γ + δ) x + γδ = 0

⇒ x^{2} – qx + 8 = 0 ……….(3)

∴ From (1), (2) and (3)

x^{4} + x^{3} – 16x^{2} – 4x + 48 = (x^{2} – px + 6) (x^{2} – qx + 8)

= x^{4} – (p + q) x^{3} + (pq + 14) x^{2} – (8p + 86q) x + 48

Comparing the like terms,

p + q = -1

8p + 6q = 4 ⇒ 4p + 3q = 2

Solving, q = -6

∴ p = -1 + 6 = 5

Substitute the value of p in eq. (2),

x^{2} – 5x + 6 = 0 ⇒ x = 2, 3

Substitute/the value of q in eq. (3),

x^{2} + 6x + 8 = 0 ⇒ x = -2, – 4

∴ The roots of the given equation are -4, -2, 2, 3

Question 2.

Solve 8x^{4} – 2x^{3} – 27x^{2} + 6x + 9 = 0 given that two roots have the same absolute value, but are opposite in signs.

Solution:

Suppose α, β, γ, δ are the roots of the equation

8x^{4} – 2x^{3} – 27x^{2} + 6x + 9 = 0

⇒ \(x^{4}-\frac{1}{4} x^{3}-\frac{27}{8} x^{2}+\frac{3}{4} x+\frac{9}{8}=0\) …………(1)

Sum of the roots = α + β + γ + δ = \(\frac{1}{4}\) and

Product of the roots = αβγδ = \(\frac{9}{8}\)

Given β = -α

⇒ α + β = 0

∴ 0 + γ + δ = \(\frac{1}{4}\)

⇒ γ + δ = \(\frac{1}{4}\)

Let αβ = p, γδ = q, so that pq = \(\frac{9}{8}\)

The equation having the roots α, β is x^{2} – (α + β)x + αβ = 0

⇒ x^{2} + p = 0 ……….(2)

The equation having the roots γ, δ is x^{2} – (γ + δ)x + γδ = 0

⇒ x^{2} – \(\frac{1}{4}\) x + q = 0 ……..(3)

From (1), (2) and (3)

\(x^{4}-\frac{1}{4} x^{3}-\frac{27}{8} x^{2}+\frac{3}{4} x+\frac{9}{8}\) = (x^{2} + p) \(\left(x^{2}-\frac{1}{4} x+q\right)\)

= \(x^{4}-\frac{1}{4} x^{3}+(p+q) x^{2}-\frac{p}{4} x+p q\)

Comparing the coefficients of x and constants

\(\frac{-p}{4}=\frac{3}{4}\) ⇒ p = -3

pq = \(\frac{9}{8}\)

⇒ q = \(\frac{9}{8} \times \frac{-1}{3}=\frac{-3}{8}\)

Substitute the value of p in eq. (2),

x^{2} – 3 = 0 ⇒ x = ±√3

Substitute the value of q in eq. (3),

\(x^{2}-\frac{1}{4} x-\frac{3}{8}=0\)

⇒ 8x^{2} – 2x – 3 = 0

⇒ (2x + 1) (4x – 3) = 6

⇒ x = \(-\frac{1}{2}, \frac{3}{4}\)

∴ The roots of the given equation are -√3, \(-\frac{1}{2}, \frac{3}{4}\), √3

Question 3.

Solve 18x^{3} + 81x^{2} + 121x + 60 = 0 given that one root is equal to half the sum of the remaining roots.

Solution:

Suppose α, β, γ are the roots of 18x^{3} + 81x^{2} + 121x + 60 = 0

sum = α + β + γ = \(\frac{-81}{18}=\frac{-9}{2}\) ……….(1)

αβ + βγ + γα = \(\frac{121}{18}\) …………(2)

αβγ = \(\frac{-60}{18}=\frac{-10}{3}\) ………(3)

∵ One root is equal to half of the sum of the remaining two,

Let α = \(\frac{1}{2}\) (β + γ)

Substitute in (1)

Question 4.

Find the condition in order that the equation ax^{4} + 4bx^{3} + 6cx^{2} + 4dx + e = 0 may have a pair of equal roots.

Solution:

Let α, α, β, β are the roots of the equation.

ax^{4} + 4bx^{3} + 6cx^{2} + 4dx + e = 0

⇒ \(x^{4}+\frac{4 b}{a} x^{2}+\frac{6 c}{a} x^{2}+\frac{4 d}{a} x+\frac{e}{a}=0\)

Sum of the roots, 2(α + β) = \(-\frac{4 b}{a}\)

⇒ α + β = \(-\frac{2 b}{a}\)

⇒ αβ = k (say)

Equation having roots α, β is x^{2} – (α + β) x + αβ = 0

Question 5.

(i) Show that x^{5} – 5x^{3} + 5x^{2} – 1 = 0 has three equal roots and find this root.

Solution:

Let f(x) = x^{5} – 5x^{3} + 5x^{2} – 1

f'(x) = 5x^{4} – 15x^{2} + 10x = 5x(x^{3} – 3x + 2)

f'(1) = 5(1) (1 – 3 + 2) = 0

f(1) = 1 – 5 + 5 – 1 = 0

x – 1 is a factor of f'(x) and f(x)

∴ 1 is a repeated root of f(x).

x^{3} + 2x^{2} – 2x – 1 = 0

⇒ 1 is a root of the above equation (∵ sum of the coefficients is zero)

∴ 1 is the required root.

(ii) Find the repeated roots of x^{5} – 3x^{4} – 5x^{3} + 27x^{2} – 32x + 12 = 0

Solution:

Let f(x) = x^{5} – 3x^{4} – 5x^{3} + 27x^{2} – 32x + 12

f'(x) = 5x^{4} – 12x^{3} – 15x^{2} + 54x – 32

f'(2) = 5(2)^{4} – 12(2)^{3} – 15(2)^{2} + 54(2) – 32

= 80 – 96 – 60 + 108 – 32

= 0

f(2) = (2)^{5} – 3(2)^{4} – 5(2)^{3} + 27(2)^{2} – 32(2) + 12

= 32 – 48 – 40 + 108 – 64 + 12

= 152 – 152

= 0

∴ x – 2 is a common factor of f'(x) and f(x)

2 is a multiple root of f(x) = 0

Let g(x) = x^{3} + x^{2} – 5x + 3

g'(x) = 3x^{2} + 2x – 5 = (3x + 5) (x – 1)

g(1) = 1 + 1 – 5 + 3 = 0

∴ x – 1 is a common factor of g'(x) and g(x)

∴ 1 is a multiple root of g(x) = 0

x + 3 = 0 ⇒ x = -3

∴ The roots are 2, 2, 1, 1, -3

Question 6.

Solve the equation 8x^{3} – 20x^{2} + 6x + 9 = 0 given that the equation has multiple roots.

Solution:

Let f(x) = 8x^{3} – 20x^{2} + 6x + 9

f'(x) = 24x^{2} – 40x + 6

= 2 (12x^{2} – 20x + 3)

= 2[12x^{2} – 18x – 2x + 3]

= 2[6x(2x – 3) – 1(2x – 3)]

= 2(2x – 3) (6x – 1)

f'(x) = 0

⇒ x = \(\frac{3}{2}\), x = \(\frac{1}{6}\)

\(f\left(\frac{3}{2}\right)=8\left(\frac{3}{2}\right)^{3}-20\left(\frac{3}{2}\right)^{2}+6\left(\frac{3}{2}\right)+9\)

= 27 – 45 + 9 + 9

= 0

Hence x – \(\frac{3}{2}\) is a factor of f(x) and f'(x)

∴ \(\frac{3}{2}\) is a multiple root of f(x) = 0

8x – 4 = 0

⇒ x = \(\frac{4}{8}=\frac{1}{2}\)

∴ The roots of the equation f(x) = 0 are \(\frac{3}{2}, \frac{3}{2}, \frac{1}{2}\)