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 x3 – 3x2 – 16x + 48 = 0, given that the sum of two roots is zero.
Solution:
Let α, β, γ are the roots of x3 – 3x2 – 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 x3 – 3x2 – 16x + 48 = 0

x2 – 16 = 0
⇒ x2 = 16
⇒ x = ±4
∴ The roots are -4, 4, 3

Question 2.
Find the condition that x3 – px2 + qx – r = 0 may have the sum of its roots zero.
Solution:
Let α, β, γ be the roots of x3 – px2 + 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 x3 – px2 + qx – r = 0
γ3 – pγ2 + qγ – r = 0
But γ = p
p3 – p(p2) + q(p) – r = 0
p3 – p2 + qp – r = 0
∴ qp = r is the required condition.

Question 3.
Given that the roots of x3 + 3px2 + 3qx + r = 0 are in
(i) A.P., show that 2p3 – 3qp + r = 0
(ii) G.P., show that p3r = q3
(iii) H.P., show that 2q3 = r(3pq – r)
Solution:
Given equation is x3 + 3px2 + 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 x3 + 3px2 + 3qx + r = 0
⇒ a3 + 3pa2 + 3qa + r = 0
But a = -p
⇒ p3 + 3p(-p)2 + 3q(-p) + r = 0
⇒ 2p3 – 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 x3 + 3px2 + 3qx + r = 0
⇒ (-r1/3)3 + 3p(-r1/3)2 + 3q(-r1/3) + r = 0
⇒ -r + 3pr2/3 – 3qr1/3 + r = 0
pr2/3 = qr1/3
⇒ pr1/3 = q
⇒ p1/3r = q is the required condition

(iii) The roots of x3 + 3px2 + 3qx + r = 0 are in H.P.
x3 + 3px2 + 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 ry3 + 3qy2 + 3py + 1 = 0 are in A.P.
ry3 + 3qy2 + 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 ry3 + 3qy2 + 3py + 1 = 0
⇒ ra3 + 3qa2 + 3pa + 1 = 0
But a = $$-\frac{q}{r}$$
⇒ $$r\left(-\frac{q}{r}\right)^{3}+3 q\left(-\frac{q}{r}\right)^{2}+3 p\left(-\frac{q}{r}\right)+1=0$$
⇒ $$\frac{-q^{3}}{r^{2}}+\frac{3 q^{3}}{r^{2}}-\frac{3 p q}{r}+1=0$$
⇒ -q3 + 3q3 – 3pqr + r2 = 0
⇒ 2q3 = r(3pq – r) is the required condition.

Question 4.
Find the condition that x3 – px2 + qx – r = 0 may have the roots in G.P.
Solution:
Let $$\frac{a}{R}$$, a, aR be the roots of x3 – px2 + qx – r = 0
The product of the roots = $$\frac{a}{R}$$ . a . aR = a3
product of the roots = r
⇒ a = r1/3
∵ a is a root of x3 – px2 + qx – r = 0
⇒ a3 – pa2 + qa – r = 0
But a = r1/3
⇒ (r1/3)3 – p(r1/3)2 + q(r1/3) – r = 0
⇒ r – p . r2/3 + q . r1/3 – r = 0
⇒ p . r2/3 = qr1/3
By cubing on both sides
⇒ p3r2 = q3r
⇒ p3r = q3 is the required condition

II.

Question 1.
Solve 9x3 – 15x2 + 7x – 1 = 0, given that two of its roots are equal.
Solution:
Suppose α, β, γ are the roots of 9x3 – 15x3 + 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
α2 + 2αγ = $$\frac{7}{9}$$
⇒ α2 + 2α ($$\frac{5}{3}$$ – 2α) = $$\frac{7}{9}$$
⇒ α2 + $$\frac{2 \alpha(5-6 \alpha)}{3}=\frac{7}{9}$$
⇒ 9α2 + 6α(5 – 6α) = 7
⇒ 9α2 + 30α – 36α2 = 7
⇒ 27α2 – 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 2x3 + 3x2 – 8x + 3 = 0 is double of another root, find the roots of the equation.
Solution:
Suppose α, β, γ are the roots of 2x3 + 3x2 – 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
⇒ 2β2 + 3βγ = -4
⇒ 2β2 + 3β($$-\frac{3}{2}$$ – 3β) = -4
⇒ 2β2 – $$\frac{3 \beta(3+6 \beta)}{2}$$ = -4
⇒ 4β2 – 9β – 18β2 = -8
⇒ 14β2 + 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 x3 – 9x2 + 14x + 24 = 0, given that two of the roots are in the ratio 3 : 2.
Solution:
Suppose α, β, γ are the roots of x3 – 9x2 + 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)
⇒ $$\frac{2}{3}$$ α2 + γ(α + β) = 14
⇒ $$\frac{2}{3} \alpha^{2}+\left(9-\frac{5 \alpha}{3}\right) \cdot \frac{5 \alpha}{3}$$ = 14
⇒ 2α2 + 5α(9 – $$\frac{5 \alpha}{3}$$) = 42
⇒ 2α2 + 5α $$\frac{(27-5 \alpha)}{3}$$ = 42
⇒ 6α2 + 135α – 25α2 = 126
⇒ 19α2 – 135α + 126 = 0
⇒ 19α2 – 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) 8x3 – 36x2 – 18x + 81 = 0
Solution:
Given the roots of 8x3 – 36x2 – 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 8x3 – 36x2 – 18x + 81 = 0

⇒ 8x2 – 24x – 54 = 0
⇒ 4x2 – 12x – 27 = 0
⇒ 4x2 – 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) x3 – 3x2 – 6x + 8 = 0
Solution:
The roots of x3 – 3x2 – 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 x3 – 3x2 – 6x + 8 = 0

⇒ x2 – 2x – 8 = 0
⇒ x2 – 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) 3x3 – 26x2 + 52x – 24 = 0
Solution:
Given equation is 3x3 – 26x2 + 52x – 24 = 0
The roots are in G.P.
Suppose $$\frac{a}{r}$$, a, ar are the roots.
Product = $$\frac{a}{r}$$ . a . ar = $$-\left(-\frac{24}{3}\right)$$
⇒ a3 = 8
⇒ a = 2
∴ (x – 2) is a factor of 3x3 – 26x2 + 52x – 24

Hint: 3 × 12 = 3 ×6 × 2 = (-18)(-2)
⇒ 3x2 – 20x + 12 = 0
⇒ 3x2 – 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) 54x3 – 39x2 – 26x + 16 = 0
Solution:
Given equation is 54x3 – 39x2 – 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)
⇒ 54x2 – 75x + 24 = 0
⇒ 18x2 – 25x + 8 = 0
⇒ 18x2 – 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) 6x3 – 11x2 + 6x – 1 = 0
Solution:
Given equation is 6x3 – 11x2 + 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 + 6y2 – y3 = 0
⇒ y3 – 6y2 + 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(a2 – d2) = 6
⇒ 2(4 – d2) = 6
⇒ 4 – d2 = 3
⇒ d2 = 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) 15x3 – 23x2 + 9x – 1 = 0
Solution:
Given equation is 15x3 – 23x2 + 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 + 9y2 – y3 = 0
⇒ y3 – 9y2 + 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(a2 – d2) = 15
⇒ 3(9 – d2) = 15
⇒ 9 – d2 = 5
⇒ d2 = 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) x4 – 6x3 + 13x2 – 24x + 36 = 0
Solution:
(i) Let f(x) = x4 – 6x3 + 13x2 – 24x + 36
⇒ f'(x) = 4x3 – 18x2 + 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)

x2 + 4 = 0
⇒ x = ±2i
∴ The roots of the given equation are 3, 3, ±2i

(ii) 3x4 + 16x3 + 24x2 – 16 = 0
Solution:
Let f(x) = 3x4 + 16x3 + 24x2 – 16
f(x) = 12x3 + 48x2 + 48x
= 12x(x2 + 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

3x2 + 4x – 4 = 0
⇒ 3x2 + 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 x4 + x3 – 16x2 – 4x + 48 = 0, given that the product of two of the roots is 6.
Solution:
Suppose α, β, γ, δ are the roots of
x4 + x3 – 16x2 – 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 x2 – (α + β) x + αβ = 0
⇒ x2 – px + 6 = 0 ………..(2)
The equation having the roots γ, δ is x2 – (γ + δ) x + γδ = 0
⇒ x2 – qx + 8 = 0 ……….(3)
∴ From (1), (2) and (3)
x4 + x3 – 16x2 – 4x + 48 = (x2 – px + 6) (x2 – qx + 8)
= x4 – (p + q) x3 + (pq + 14) x2 – (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),
x2 – 5x + 6 = 0 ⇒ x = 2, 3
Substitute/the value of q in eq. (3),
x2 + 6x + 8 = 0 ⇒ x = -2, – 4
∴ The roots of the given equation are -4, -2, 2, 3

Question 2.
Solve 8x4 – 2x3 – 27x2 + 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
8x4 – 2x3 – 27x2 + 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 x2 – (α + β)x + αβ = 0
⇒ x2 + p = 0 ……….(2)
The equation having the roots γ, δ is x2 – (γ + δ)x + γδ = 0
⇒ x2 – $$\frac{1}{4}$$ x + q = 0 ……..(3)
From (1), (2) and (3)
$$x^{4}-\frac{1}{4} x^{3}-\frac{27}{8} x^{2}+\frac{3}{4} x+\frac{9}{8}$$ = (x2 + 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),
x2 – 3 = 0 ⇒ x = ±√3
Substitute the value of q in eq. (3),
$$x^{2}-\frac{1}{4} x-\frac{3}{8}=0$$
⇒ 8x2 – 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 18x3 + 81x2 + 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 18x3 + 81x2 + 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 ax4 + 4bx3 + 6cx2 + 4dx + e = 0 may have a pair of equal roots.
Solution:
Let α, α, β, β are the roots of the equation.
ax4 + 4bx3 + 6cx2 + 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 x2 – (α + β) x + αβ = 0

Question 5.
(i) Show that x5 – 5x3 + 5x2 – 1 = 0 has three equal roots and find this root.
Solution:
Let f(x) = x5 – 5x3 + 5x2 – 1
f'(x) = 5x4 – 15x2 + 10x = 5x(x3 – 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).

x3 + 2x2 – 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 x5 – 3x4 – 5x3 + 27x2 – 32x + 12 = 0
Solution:
Let f(x) = x5 – 3x4 – 5x3 + 27x2 – 32x + 12
f'(x) = 5x4 – 12x3 – 15x2 + 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) = x3 + x2 – 5x + 3
g'(x) = 3x2 + 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 8x3 – 20x2 + 6x + 9 = 0 given that the equation has multiple roots.
Solution:
Let f(x) = 8x3 – 20x2 + 6x + 9
f'(x) = 24x2 – 40x + 6
= 2 (12x2 – 20x + 3)
= 2[12x2 – 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 $$\frac{3}{2}, \frac{3}{2}, \frac{1}{2}$$