Students must practice this TS Intermediate Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) to find a better approach to solving the problems.

## TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 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 be the roots of

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

Given that the sum of two roots is zero.

Let α + β = 0 …………(2)

But from (1) we have α + β + γ = 3

⇒ γ = 3

Hence αβγ = – 48

⇒ αβ = – 16

We know that,

(α + β)^{2} – (α – β)^{2} = 4αβ

⇒ (α – β)^{2} = 64

⇒ α – β = ± 8.

i) When α – β = 8

⇒ 2α = 8 (∵ from (2))

⇒ α = 4

∴ β = – 4.

ii) When α – β = – 8

⇒ 2α = – 8 (∵ from (2))

⇒ α = – 4

∴ β = 4.

The roots of given equation are 4, – 4 and 3.

Question 2.

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

Solution:

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

Let α, β, γ be the roots.

∴ α + β + γ = p …………..(1)

Given sum of two of its roots is zero.

∴ (1) ⇒ α + 0 = p

i. e„ α = p

Substituting in given equation, we get p^{3} – p^{3} + pq – r = 0

⇒ pq – r = 0

⇒ Pq = r.

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 cubic equation is

x^{3} + 3px^{2} + 3qx + r = 0 ……………..(1)

Let α, β, γ be its roots.

i) When the roots are in A.P. :

i.e., 2β = α + γ

from (1) α + β + γ = – 3p

⇒ (α + γ) + P = – 3p

⇒ 2β + β = – 3p

⇒ 3β = – 3p

⇒ β = – P

Substituting in given equation, we get

– p^{3} + 3p^{3} – 3pq + r = 0

⇒ 2p^{3} – 3pq + r = 0.

ii) When the roots are in G.P. :

∴ β^{2} = αγ

from (1) αβγ = – r

⇒ β^{3} = – r

⇒ β = – r^{1/3}

Substituting in (1), we get

(- r^{1/3})^{3} + 3pr^{2/3} – 3qr^{1/3} + r = 0

⇒ 3pr^{2/3} = 3qr^{1/3}

⇒ P^{3}r = q^{3}.

iii) When the roots are in H.P. :

β = \(\frac{2 \alpha \gamma}{\alpha+\gamma}\)

⇒ β^{2} = \(\frac{2 \alpha \beta \gamma}{(\alpha+\beta+\gamma)-\beta}\)

⇒ β^{2} = \(\frac{-2 r}{-3 p-\beta}\)

⇒ β^{2} + 3pβ^{2} = 2r (∵ β is a root of (1))

⇒ – 3qβ – r = 2r

β = – \(\frac{r}{q}\)

Substituting in (1) we get

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

⇒ 2q^{3} = r(3pq – r).

Question 4.

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

Solution:

Let α, β, γ be the roots of

x^{3} – px^{2} + qx – r = 0

If α, β, γ are in G.P., then

β^{2} = αγ

(1) ⇒ αβγ = r

⇒ β^{3} = r

β = r^{1/3}

Substituting in (1), we get

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

⇒ pr^{2/3} = qr^{1/3}

⇒ p^{3}r^{2} = q^{3}r

⇒ q^{3} = p^{3}r.

II.

Question 1.

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

Solution:

Given cubic equation is

9x^{3} – 15x^{2} – 7x – 1 = 0 ……………. (1)

Suppose α, β, γ are the roots of (1)

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

αβ + βγ + γα = \(\frac{7}{9}\)

αβγ = \(\frac{1}{9}\)

According to the problem, α = β (∵ two of its roots are equal)

∴ 2α + γ = \(\frac{5}{3}\)

⇒ γ = \(\frac{5}{3}\) – 2α

Also, α^{2} + 2αγ = \(\frac{7}{9}\)

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

⇒ 27α^{2} – 30α + 7 = 0

⇒ (3α – 1) (9α – 7) = 0

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

Case (i) :

when α = \(\frac{1}{3}\)

γ = \(\frac{5}{3}\) – 2α

= \(\frac{5}{3}\) – \(\frac{2}{3}\) = 1

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

Case – (ii):

When α = \(\frac{7}{9}\)

γ = \(\frac{5}{3}\) – 2α

= \(\frac{5}{3}-\frac{14}{9}\) = \(\frac{1}{9}\)

Which is impossible as

αβγ = \(\frac{7}{9} \cdot \frac{7}{9} \cdot \frac{1}{9}\) ≠ \(\frac{1}{9}\)

∴ 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 the other root, find the roots of equation.

Solution:

Given cubic equation is

2x^{3} + 3x^{2} – 8x + 3= 0 ……………..(1)

Suppose α, β, γ are the roots of (1).

∴ α + β + γ = \(\frac{-3}{2}\)

αβ + βγ + γα = \(\frac{-8}{2}\) = – 4 …………….(2)

αβγ = \(\frac{-3}{2}\)

Given one root is double the other.

3α + γ = \(\frac{-3}{2}\)

⇒ γ = \(\frac{-3}{2}\) – 3α

Also from (2):

2α^{2} – 3α (\(\frac{3}{2}\) + 3α) = – 4

14α^{2} + 9α – 8 = 0

(2α – 1) (7α + 8) = 0

α = \(\frac{1}{2}\) or α = \(\frac{-8}{7}\).

Case (i):

When α = \(\frac{1}{2}\)

β = 2α = 2 (\(\frac{1}{2}\)) = 1

γ = \(\frac{-3}{2}\) – 3α

= \(\frac{-3}{2} \frac{-3}{2}\) = – 3.

∴ α = \(\frac{1}{2}\), β = 1 and γ = – 3

satisfies αβγ = \(\frac{-3}{2}\)

∴ The roots are \(\frac{1}{2}\), 1, – 3.

Case (ii):

When α = \(\frac{-8}{7}\)

β = 2α = \(\frac{-16}{7}\)

γ = \(\frac{-3}{2}\) – 3α

= \(\frac{-3}{2}+\frac{48}{7}=\frac{75}{14}\)

But α = \(\frac{-8}{7}\), β = \(\frac{-16}{7}\) and γ = \(\frac{75}{14}\) do not satisfy αβγ = \(\frac{-3}{2}\).

Hence the roots of given equation are \(\frac{1}{2}\), 1, – 3.

Question 3.

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

Solution:

Given cubic equation is

x^{3} – 9x^{2} + 14x + 24 = 0 ……….(1)

Let α, β, γ be the roots of (1)

∴ α + β + γ = 9, αβ + βγ + γα = 14, αβγ = – 24 ……………..(2)

Given two roots are in the ratio 3 : 2,

let α : β = 3 : 2

⇒ β = \(\frac{2 \alpha}{3}\)

Now from (2) \(\frac{5 \alpha}{3}\) + γ = 9

⇒ γ = 9 – \(\frac{5 \alpha}{3}\)

Also, \(\frac{2}{3}\) α^{2} + (9 – \(\frac{5 \alpha}{3}\)) \(\frac{5 \alpha}{3}\) = 14

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

⇒ 19α^{2} – 135α + 126 = 0

⇒ (19α – 21) (α – 6) = 0

⇒ α = \(\frac{21}{19}\) or α = 6.

Case (i):

When α = \(\frac{21}{19}\)

β = \(\frac{2}{3}(\alpha)=\frac{2}{3}\left(\frac{21}{19}\right)=\frac{14}{19}\)

γ = \(9-\frac{5 \alpha}{3}=9-\frac{5}{3}\left(\frac{21}{19}\right)=\frac{136}{19}\)

These values do not satisfy αβγ = – 24.

Case – (ii) :

When α = 6

β = \(\frac{2}{3}(\alpha)=\frac{2}{3}(6)\) = 4

γ = \(9-\frac{5 \alpha}{3}=9-\frac{5}{3}(6)\) = – 1

These values satisfy αβγ = – 24.

∴ The roots of given equation 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

ii) x^{3} – 3x^{2} – 6x + 8 = 0

Solution:

i) Given cubic equation is

8x^{3} – 36x^{2} – 18x + 81 = 0 …………….(1)

Given the roots are in A.P.

∴ α – d, α, α + d be the roots.

∴ Sum of the roots 3α = \(\frac{36}{8}\)

⇒ α = \(\frac{3}{2}\)

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

By synthetic division,

8x^{3} – 36x^{2} – 18x + 81 = (x – \(\frac{3}{2}\)) (8x^{2} – 24x – 54)

∴ Equation (1)

⇒ (x – \(\frac{3}{2}\)) (8x^{2} – 24x – 54) = 0

⇒ (x – \(\frac{3}{2}\)) (2x + 3) (2x – 9) = 0

⇒ x = – \(\frac{3}{2}\) or x = \(\frac{3}{2}\) or x = \(\frac{9}{2}\)

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

ii) Given roots of cubic equation

x^{3} – 3x^{2} – 6x + 8 = 0 ……………(1) are in G.P.

Let α – d, α, α + d be the roots.

∴ Sum of the roots 3α = 3

⇒ α = 1

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

By synthetic division,

x^{3} – 3x^{2} – 6x + 8 = (x – 1) (x^{2} – 2x – 8)

∴ Equation (1)

⇒ (x – 1) (x^{2} – 2x – 8) = 0

⇒ (x – 1) (x – 4) (x + 2) = 0

∴ x = 1 or x = 4 or x = – 2.

∴ The roots are – 2, 1, 4.

Question 5.

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

i) 3x^{3} – 26x^{2} + 52x – 24= 0

ii) 54x^{3} – 39x^{2} – 26x + 16 = 0

Solution:

i) Given roots of cubic equation

3x^{3} – 26x^{2} + 52x – 24 = 0 ……………. (1) are in G.P.

Let \(\frac{\alpha}{r}\), α, αr be the roots.

∴ Product of the roots α^{3} = \(\frac{24}{3}\) = 8

⇒ α = 2

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

By synthetic division,

3x^{3} – 26x^{2} + 52x – 24 = (x – 2) (3x^{2} – 20x + 12)

∴ Equation (1) ⇒ (x – 2) (3x^{2} – 20x + 12) = 0

⇒ (x – 2) (3x – 2) (x – 6) = 0

∴ x = 2 or x = \(\frac{2}{3}\) or x = 6

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

ii) Given roots of cubic equation.

54x^{3} – 39x^{2} – 26x + 16 = 0 (1) are in GP.

Let \(\frac{\alpha}{r}\), α, αr be the roots.

∴ Product of the roots α^{3} = \(\frac{-16}{54}\)

⇒ α^{3} = \(\frac{-2}{3}\)

∴ (x + \(\frac{2}{3}\)) is a factor 54x^{3} – 39x^{2} – 26x + 16

By synthetic division,

54x^{3} – 39x^{2} – 26x + 16 = (x + \(\frac{2}{3}\)) (54x^{2} – 75x + 24)

∴ Equation (1),

(x + \(\frac{2}{3}\)) (54x^{2} – 75x + 24) = 0

(x + \(\frac{2}{3}\)) (18x^{2} – 25x + 8) = 0

(x + \(\frac{2}{3}\)) (9x – 8)(2x – 1) = 0

∴ x = – \(\frac{2}{3}\) or x = \(\frac{8}{9}\) or x = \(\frac{1}{2}\)

∴ 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

ii) 15x^{3} – 23x^{2} – 9x – 1 = 0

Solution:

i) Given cubic equation is

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

Put y = \(\frac{1}{x}\)

∴ (1) ⇒ \(\frac{6}{y^3}-\frac{11}{y^2}+\frac{6}{y}\) – 1 = 0

⇒ y3 – 6y2 + 11y – 6 = 0 ………… (2)

Given roots of (1) are in H.P.

⇒ Roots of (2) are in AP.

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

∴ Sum of the roots, 3a = 6

⇒ α = 2

∴ (x – 2) is a factor of y^{3} – 6y^{2} + 11y – 6

By synthetic division

∴ y^{3} – 6y^{2} + 11y – 6 = (y – 2) (y^{2} – 4y 3)

∴ Equation (2) = (y – 2) (y^{2} – 4y + 3) = 0

⇒ (y – 2) (y – 3) (y – 1) = 0

∴ y = 1 or y = 2 ory = – 3

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

Hence the roots of (1) are 1, \(\frac{1}{2}\), \(\frac{1}{3}\).

ii) Given cubic equation is

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

put y = \(\frac{1}{x}\)

∴ (1) ⇒ y^{3} – 9y + 23y^{2} – 15 = 0 ………..(2)

Given roots of (1) are in 1-LP.

⇒ Roots of (2) are in A.P.

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

∴ Sum of the roots, 3α = 9

⇒ α = 3

∴ (y – 3) is a factor of y^{3} – 9y + 23y^{2} – 15.

By synthetic division,

∴ y^{3} – 9y + 23y^{2} – 15 = (y – 3) (y^{2} – 6y + 5)

∴ Equation (2) = (y – 3) (y2 – 6y + 5) = 0

⇒ (y – 3) (y – 1) (y – 5) = 0

∴ y = 1 or y = 3 or y = 5

∴ The roots of (2) are 1, 3, 5.

Hence the roots of (2) 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

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

Solution:

i) Given equation,

x^{4} – 6x^{3} + 13x^{2} – 24x + 36 = 0 …………..(1)

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

f’(x) = 4x^{3} – 18x^{2} + 26x – 24

= 2 (2x^{3} – 9x^{2} + 13x – 12)

f’(3) = 2(54 – 81 + 39 – 12)

⇒ f'(3) = o

Now

f(3) = 81 – 162 + 117 – 72 + 36

= f(3) = 0

∴ (x – 3) is a factor of f(x) and f’(x).

∴ 3 is the repeated root of f(x) = 0.

Now we divide f(x) by (x – 3) by using synthetic division.

∴ f(x) = (x – 3) (x – 3) (x2 + 4)

∴ Equation (1)

⇒ f(x) = 0

⇒ (x – 3) (x – 3) (x^{2} + 4) = 0

∴ x = 3 or x^{2} + 4 = 0

⇒ x = ±2i

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

ii) Given equation is

3x^{4} + 16x^{3} + 24x^{2} – 16 = 0 …………..(1)

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

⇒ f'(x) = 12x^{3} + 48x^{2} + 48x

= 12 (x^{3} + 4x^{2} + 4x)

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

⇒ f’ (- 2) = 0

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

∴ (x + 2) is a factor of f(x) and f'(x).

∴ – 2 is a repeated root of f(x) = 0.

Now we divide f(x) by (x + 2) using synthetic division.

∴ f(x) = (x + 2) (x + 2) (3x^{2} + 4x – 4)

Equation (1)

⇒ f(x) = 0

⇒ (x + 2) (x + 2) (3x^{2} + 4x – 4) = 0

⇒ (x + 2) (x + 2) (3x – 2) (x + 2) = 0

⇒ x = – 2 or x = \(\frac{2}{3}\)

∴ The roots of 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:

Given equation is

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

Let α, β, γ, δ be the roots

∴ x^{4} + x^{3} – 16x^{2} – 4x + 48 = (x + α) (x – β) (x – γ) (x – δ) ……….(2)

∴ Sum of the roots α + β + γ + δ = – 1

and product of roots ⇒ αβγδ = 48 …………..(3)

Given product of two roots = 6

Let αβ = 6

∴ γδ = \(\frac{48}{\alpha \beta}\)

γδ = 8

Let α + β = a and γ + δ = b

Now (2)

⇒ x^{4} + x^{3} – 16x^{2} – 4x + 48 = (x^{2} – (α + β) x + αβ) (x^{2} – (γ + δ) x + γδ)

⇒ x^{4} + x^{3} – 16x^{2} – 4x + 48 = (x^{2} – ax + 6) (x^{2} – bx + 8)

Comparing like terms.

we get, a + b = – 1 and

8a – 6b = 5 ……………..(4)

⇒ 4a + 3b = 2 (5)

(5) ⇒ 4a + 3 (- 1 – a) = 2 (∵ from (4))

⇒ a = 5

∴ b = – 6

∴ x^{4} + x^{3} – 16x^{2} – 4x + 48 = (x^{2} – 5x + 6) (x^{2} + 6x + 8)

= (x – 2) (x – 3) (x + 2) (x + 4)

∴ Equation (1),

⇒ (x – 2) (x – 3) (x + 2) (x + 4) = 0

∴ x = – 4; x = – 2 or x = 2 or x = 3

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

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 sign.

Solution:

Given equation is

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)

Let α, β, γ, δ be the roots of (1)

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

and product of roots αβγδ = \(\frac{9}{8}\)

But given two roots have same absolute value but are opposite sign.

Let α = – β

⇒ α + β = 0

∴ γ + δ = \(\frac{-1}{4}\)

Let αβ = a and γδ = b

Now(x – α) (x – β) = x^{2} – (α + β)x + αβ

⇒ (x – α) (x – β) = x^{2} + a …………(2)

Also (x – γ) (x – δ) = x^{2} – (γ + δ)x + γδ

= (x – γ) (x – δ) = x^{2} – \(\frac{1}{4}\) x + b ………….(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} + a) (x^{2} – \(\frac{1}{4}\) x + b)

Comparing like terms,

\(\frac{3}{4}=\frac{-a}{4}\) and ab = \(\frac{9}{8}\)

a = – 3

∴ b = \(\frac{9}{8(-3)}\)

b = \(\frac{-3}{8}\)

∴ (2) ⇒ (x – α) (x – β) = x^{2} – 3

& (3) ⇒ (x – γ) (x – δ) = (x^{2} – \(\frac{1}{4}\) x + \(\frac{3}{8}\))

⇒ \(\frac{1}{8}\) (8x^{2} – 2x – 3)

⇒ (x – γ) (x – δ) = \(\frac{1}{8}\) (2x + 1) (4x – 3)

(x – γ) (x – δ) = (x + \(\frac{1}{2}\)) (x – \(\frac{3}{4}\))

∴ Equation (1)

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

⇒ x = ± √3 or x = – \(\frac{1}{2}\) or x = \(\frac{3}{4}\)

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

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:

Given equation is

18x^{3} + 81x^{2} + 121x + 60 = 0 ……………(1)

Let α, β, γ, δ be the roots

∴ Sum of roots, α + β + γ = \(\frac{-81}{18}=\frac{-9}{2}\)

αβ + βγ + γδ = \(\frac{121}{18}\)

and product of roots αβγ = \(\)

given one root is equal to halt of the sum of the remaining roots.

∴ Let α = \(\frac{\beta+\gamma}{2}\)

∴ α + 2α = \(\frac{-9}{2}\)

⇒ 3α = \(\frac{-9}{2}\)

⇒ α = \(\frac{-9}{2}\)

∴ x + \(\frac{3}{2}\) is a factor of 18x^{3} + 81x^{2} + 121x + 60.

By synthetic division,

∴ 18x^{3} + 81x^{2} + 121x + 60 = (x + \(\frac{3}{2}\)) (18x^{2} + 54x + 40)

= (x + \(\frac{3}{2}\)) (9x^{2} + 27x + 60)

∴ 18x^{3} + 81x^{2} + 121x + 60 = 2 (x + \(\frac{3}{2}\)) (3x + 4) (3x + 5)

∴ Equation (1),

⇒ 2 (x + \(\frac{3}{2}\)) (3x + 4) (3x + 5) = 0

∴ x = \(\frac{-3}{2}\) or x = \(\frac{-4}{3}\) or x = \(\frac{-5}{3}\).

∴ The roots of given equation are \(\frac{-3}{2}\), \(\frac{-4}{3}\), \(\frac{-5}{3}\).

Question 4.

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

Solution:

Given equation is

ax^{4} + 4bx^{3} + 6cx^{2} + 4dx + e = 0 ………………..(1)

Given (1) has two pairs of equal roots.

∴ Let α, α, β, β be the root of (1).

(1) ⇒ x^{4} + \(\frac{4 b}{a} x^3+\frac{6 c}{a} x^2+\frac{4 d}{a} x+\frac{e}{a}\) = 0

3abc = 2b^{3} + a^{2}d and ad^{2} = eb^{2}.

∴ The required conditions are 2b^{3} + a^{2}d = 3abc and ad^{2} = eb^{2}.

Question 5.

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

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

Solution:

i) Given equation is x^{5} – 5x^{3} + 5x^{2} – 1 = 0

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

f’(x) = 5x^{4} – 15x^{2} + 10x

f”(x) = 20x^{3} – 30x + 10

f”(1) = 20 – 30 + 10 = 0

Similarly, f’(1) = 0 and f(1) = 0

∴ (x – 1) is a factor of f”(x), f’(x) & f(x).

Thus f(x) = 0 has three equal roots and it is ‘1’.

ii) Given equation is

x^{5} – 3x^{4} – 5x^{3} + 27x^{2} – 32x + 12 = 0 …………(1)

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

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

f’(1) = 5 – 12 – 15 + 54 – 32 = 0

Similarly f'(1) = 0 and f(1) = 0

∴ (x – 1) is a factor of f”(x), f'(x) & f(x).

Thus f(x) = 0 has three equal roots and it is ‘1’.

By synthetic division,

∴ f(x) = (x – 1)^{2} (x^{3} – x^{2} – 8x + 12)

Let g(x) = x^{3} – x^{2} – 8x + 12

g’(x) = 3x^{2} – 2x – 8

g’(2) = 3(4) – 2(2) – 8 = 0

and g(2) = 2^{3} – 2^{2} – 8(2) + 12 = 0.

∴ (x – 2) is a factor of g(x) and g’(x).

∴ 2 is a multiple root of g(x) = 0.

By synthetic division,

∴ g(x) = (x – 2)^{2} (x + 3)

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

The roots of given equation are 1, 1, 2, 2, 3.

Hence repeated roots are 1 and 2.

Question 6.

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

Solution:

Given equation is 8x^{3} – 20x^{2} + 6x + 9 = 0 …………..(1)

Let f(x) = 8x^{3} – 20x^{2} + 6x + 9

f’(x) = 24x^{2} – 40x + 6

= 2 (12x^{2} – 20x + 3)

= 2 (2x – 3) (6x – 1)

\(f\left(\frac{3}{2}\right)=8\left(\frac{27}{8}\right)-20\left(\frac{9}{4}\right)+6\left(\frac{3}{2}\right)+9\)

= 27 – 45 + 9 + 9 = 0

∴ f(\(\frac{3}{2}\)) = 0

∴ f(x) and f'(x) has a common factor ‘2x – 3’.

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

By synthetic division,

∴ 8x^{3} – 20x^{2} + 6x + 9 = 0

(x – \(\frac{3}{2}\))^{2} (8x + 4) = 0

⇒ x = \(\frac{3}{2}\) or x = \(\frac{-1}{2}\)

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