Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Theory of Equations Important Questions Long Answer Type to help strengthen their preparations for exams.

## TS Inter Second Year Maths 2A Theory of Equations Important Questions Long Answer Type

Question 1.

If the roots of the equation x^{3} + 3px^{2} + 3qx + r = 0 are in AP then show that 2p^{3} – 3pq + r = 0. [May ’08]

Solution:

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

Since the roots are in A.P then they must be of the form a – d, a, a + d.

Now, s_{1} = a – d + a + a + d = \(\frac{-3 p}{1}\)

3a = – 3p

a = – p

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

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

⇒ – p^{3} + 3p^{3} – 3pq + r = 0

⇒ 2p^{3} – 3pq + r = 0.

Question 2.

If the roots of the equation x^{3} + 3px^{2} + 3qx + r = 0 are in G.P then show that p^{3}r = q^{3}. [March ‘03]

Solution:

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

Since the roots are in G.P then they must be of the form \(\frac{a}{r}\), a, ar

Now, s_{3} = \(\frac{a}{r}\) . a . ar

= \(\frac{-\mathrm{r}}{1}\) = – r

⇒ a^{3} = – r

⇒ a = (- r)^{1/3}

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

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

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

⇒ 3p(- r)^{2/3} = – 3q(- r)^{1/3}

Cubing on both sides,

⇒ p^{3} (- r)^{2} = – q^{3}(- r)

⇒ p^{3} r^{2} = q^{3}r

⇒ p^{3}r = q^{3}

Question 3.

If the roots of the equation x^{3} + 3px^{2} + 3q + r = 0 are in H.P then show that 2q^{3} = r (3pq – r).

Solution:

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

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

The roots of \(\left(\frac{1}{x}\right)^3+3 p\left(\frac{1}{x}\right)^2+3 q\left(\frac{1}{x}\right)\) + r = 0 are in A.p.

\(\frac{1}{x^3}+\frac{3 p}{x^2}+\frac{3 q}{x}\) + r = 0 are in A.P

⇒ 1 + 3px + 3qx^{2} + rx^{3} = o

⇒ rx^{3} + 3qx^{2} + 3px + 1 = 0 ……………..(1)

Let the roots of (1) be a – d, a, a + d

s_{1} = a – d + a + a + d = \(\frac{-3 q}{r}\)

⇒ 3a = \(\frac{-3 q}{r}\)

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

Since a is a root of (1)

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

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

⇒ 2q^{3} = r (3pq – r). +

Hence proved.

Question 4.

Solve 4x^{3} – 24x^{2} + 23x + 18 = 0, given that the roots of this equation are in arithmetic

progression. [Mar. ’14, May ’06, ’01]

Solution:

Given equation is 4x^{3} – 24x^{2} + 23x + 18 = 0

Given that the roots of this equation are in A.P.

Since the roots are in A.P, they must be of the form, a – d, a, a + d.

Now, a – d + a + a + d = \(-\frac{-24}{4}\) = 6

⇒ 3a = 6

⇒ a = 2

(a – d) a (a + d) = \(\frac{-18}{4}\)

(a^{2} – d^{2})a = \(\frac{-9}{2}\)

(4 – d^{2})2 = \(\frac{-9}{2}\)

4 – d^{2} = \(\frac{-9}{4}\)

d^{2} = 4 + \(\frac{9}{4}\) = \(\frac{25}{4}\)

d = ± \(\frac{5}{2}\)

Case – I:

If a = 2, d = \(\frac{5}{2}\)

The roots of given equation are – \(\frac{-1}{2}\), 2, \(\frac{9}{2}\).

Case-2:

If a = 2, d = \(\frac{-5}{2}\)

The roots of given equation are \(\frac{9}{2}\), 2, \(\frac{-1}{2}\).

Question 5.

Solve x^{3} – 3x^{2} – 6x + 8 = 0, given that the : roots are in A.P. [March ’07]

Solution:

Given equation is x^{3} – 3x^{2} – 6x + 8 = 0

Since, the roots are in A.P., thus they must; be of the form a – d, a, a + d.

Now, a – d + a + a + d = \(\frac{-(-3)}{1}\) = 3

⇒ 3a = 3

⇒ a = 1

⇒ (a – d) a (a + d) = \(\frac{-8}{1}\)

a^{2} – d^{2} = – 8

1 – d^{2} = – 8

d = 9

⇒ d = ±3

Case – 1 : If d = 3; a = 1

The roots of the given equation are; a – d, a, a + d = – 2, 1, 4.

Case – 2: If d = – 3; – a = 1;

The roots of the given equation are (a – d) a, (a + d) = 4, 1, – 2 .

Question 6.

Solve x^{3} – 7x^{2} + 14x – 8 = 0, given that the roots are in geometric progression. [May ’07]

Solution:

Given equation is x^{3} – 7x^{2} + 14x – 8 = 0

Given that the roots are in G.P then they may be of the form \(\frac{a}{r}\), a, ar

\(\frac{a}{r}\) + a + ar = a [\(\frac{1}{r}\) + 1 + r] = \(\frac{-(-7)}{1}\)

⇒ a[\(\frac{1}{r}\) + 1 + r] = 7 …………..(1)

\(\frac{a}{r}\) . a . ar = 1

⇒ a^{3} = 8

⇒ a = 2

(1) ⇒ 2\(\left(\frac{r^2+r+1}{r}\right)\) = 7

⇒ 2r^{2} – 5r + 2 = 0

⇒ 2r^{2} – 4r – r + 2 = 0

⇒ 2r (r – 2) – 1 (r – 2) = 0

⇒ (2r – 1) (r – 2) = 0

⇒ r = \(\frac{1}{2}\) (or) r = 2

Case – 1:

If a = 2, r = \(\frac{1}{2}\)

The roots of given equation are \(\frac{a}{r}\), a, ar = 4, 2, 1

Case – 2 :

If a = 2, r = 2

The roots of given equation are \(\frac{a}{r}\), a, ar = 1, 2, 4.

Question 7.

Solve the equation 15x^{3} – 23x^{2} + 9x – 1 = 0, the roots being in H.P. [May ’02]

Solution:

Giveti equation is 15x^{3} – 23x^{2} + 9x – 1 = 0

The roots of 15x^{3} – 23x^{2} + 9x – 1 = 0 are in H.P

The roots of

\(15 \cdot \frac{1}{x^3}-23 \cdot \frac{1}{x^2}+9 \cdot \frac{1}{x}-1\) = 0 are in A.P.

15 – 23x + 9x^{2} – x^{3} = 0

⇒ x^{3} – 9x^{2} + 23x – 15 = 0

Let the roots be a – d, a, a + d

a – d + a + a + d = \(\frac{-(-9)}{1}\) = 9

⇒ 3a = 9

⇒ a = 3

(a – d) a (a + d) = \(\frac{-(-15)}{1}\) = 15

⇒ a(a^{2} – d^{2}) = 15

⇒ 3(9 – d^{2}) = 15

9 – d^{2} = 5

d^{2} = 4

d = 21

The roots of x^{3} – 9x^{2} + 23x – 15 = 0 are a – d, a, a + d =3 – 2, 3, 3 + 2 = 1, 3, 5

The roots of 15x^{3} – 23x^{2} + 9x – 1 = 0 are \(\frac{1}{1}, \frac{1}{3}, \frac{1}{5}\) = 1, \(\frac{1}{3}, \frac{1}{5}\).

Question 8.

Solve 18x^{3} + 81x^{2} + 121x + 60 = 0 given that one root is equal to half the sum of the remaining roots. [May ’11, March ‘05] [AP – Mar.2019]

Solution:

Given that one root is equal to half of the sum of the remaining roots.

Given equation is 18x^{3} + 81x^{2} + 121x + 60 = 0

i.e., the roots are in A.P.

Let a – d, a, a + d be the roots of the given equation.

∴ a – d + a + a + d = \(\frac{-81}{18}\)

3a = \(\frac{-81}{18}\)

a = \(\frac{-3}{2}\)

(a – d) a (a + d) = \(\frac{-60}{18}\)

a(a – d) = \(\frac{-81}{18}\)

\(\frac{-3}{2}\left(\frac{9}{4}-d^2\right)=\frac{-60}{18}\)

d^{2} = \(\frac{9}{4}-\frac{20}{9}=\frac{81-80}{36}=\frac{1}{36}\)

d = \(\frac{1}{6}\)

∴ The roots of the given equation are a – d, a, a + d = \(\frac{-3}{2}-\frac{1}{6}, \frac{-3}{2}, \frac{-3}{2}+\frac{1}{6}\)

= \(\frac{-5}{3}, \frac{-3}{2}, \frac{-4}{3}\)

Question 9.

Given that one root of 2x^{3} + 3x^{2} – 8x + 3 = 0 is double the other root, find the roots of the equation. [May ’03]

Solution:

Given equation is 2x^{3} + 3x^{2} – 8x + 3 = 0

Now, α + β + γ = \(\frac{-3}{2}\) …………..(1)

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

αβγ = \(\frac{-3}{2}\) ……………..(3)

Given one root is double the other then

α = 2β

(1) ⇒ 2β + β + γ = \(\frac{-3}{2}\)

⇒ 3β + γ = \(\frac{-3}{2}\)

⇒ γ = \(\frac{-3}{2}\) – 3β …………….(4)

(2) ⇒ 2β . β + βγ + γ . 2β = – 4

2β^{2} + βγ + 2βγ = – 4

2β^{2} + 3βγ = – 4 ……………..(5)

From (4) & (5)

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

2β^{2} – \(\frac{9 \beta}{2}\) – 9β^{2}=

⇒ \(\frac{4 \beta^2-9 \beta-18 \beta^2}{2}\) = – 4

⇒ – 14β^{2} – 9β = – 8

⇒ 14β^{2} + 9β – 8 = 0

⇒ 14β^{2} + 16β – 7β – 8 = 0

⇒ (2β – 1) (7β + 8) = 0

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

β = \(\frac{-8}{7}\) does not satisfy the given equation.

β = \(\frac{1}{2}\)

⇒ α = 2 . \(\frac{1}{2}\) = 1

Substitute the value of β in equation (4)

⇒ γ = \(\frac{-3}{2}\) – 3 . \(\frac{1}{2}\)

= \(\frac{-3-3}{2}=\frac{-6}{2}\) = – 3.

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

Question 10.

Solve x^{4} + x^{3} – 16x^{2} – 4x + 48 = 0, given that the product of two of the roots is 6. [TS – Mar. 2019; May ’12. ’09]

Solution:

Given equation is x^{4} + x^{3} – 16x^{2} – 4x + 48 = 0

Let α, β, γ, δ are the roots of the given equation.

Then αβγδ = \(\frac{48}{1}\) = 48

Given that the product of the two roots of the given equation is 6.

Suppose that αβ = 6

Now, 6γδ = 48

γδ = 8

Let α + β = p, γ + δ = q

The quadratic equation whose roots are α, β is

x^{2} – x(α + β) + αβ = 0

x^{2} – px + 6 = 0 ………….(1)

The quadratic equation whose roots are γ, δ is

x^{2} – x (γ + δ) + γδ = 0

⇒ x^{2} – qx + 8 = 0 ……………(2)

∴ x^{4} + x^{3} – 16x^{2} – 4x + 48

= (x^{2} – px + 6) (x^{2} – qx + 8)

= x^{4} – qx^{3} + 8x^{2} – px^{3} + pqx^{2} – 8px + 6x^{2} – 6qx + 48

= x^{4} + (- p – q)x^{3} + (pq+ 14)x^{2} + (- 8p – 6q) x + 48

Now comparing coefficients of x^{3} on both sides, we get

⇒ – p – q = 1

⇒ p + q = – 1 ………………(3)

Now comparing coefficients of x on both sides, we get

– 8p – 6q = – 4

⇒ 4p + 3q = 2 ……………….(4)

Solve (3) and (4);

⇒ p = \(\frac{-5}{-1}\)

⇒ q = \(\frac{6}{-1}\)

⇒ p = 5, q = – 6

Substitute ‘p’ value in (1),

x^{2} – 5x + 6 = 0

⇒ x^{2} – 3x – 2x + 6 = 0

⇒ x (x – 3) – 2 (x – 3) = 0

⇒ (x – 2) (x – 3) = 0

⇒ x = 2, x = 3

Substitute ‘q’ value in (2),

(2) ⇒ x^{2} + 6x + 8 = 0

⇒ x^{2} + 4x . 2x + 8 = 0

⇒ (x + 2) (x +4) = 0

⇒ x = – 4, x = – 2

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

Question 11.

Given that two roots of 4×3+20×2-23x+6=O are equal, find all the roots of the given equation.

Solution:

Given equation is 4x^{3} + 20x^{2} – 23x + 6 = 0

Let α, β, γ be the roots of given equation.

Then s_{1} = α + β + γ = \(\frac{-(20)}{4}\)

⇒ α + β + γ = – 5 …………….(1)

⇒ αβ + βγ + γα = \(\frac{-23}{4}\) ……………(2)

⇒ αβγ = \(\frac{-6}{4}=\frac{-3}{2}\) ………….(3)

Given that two roots are equal

⇒ β = α

(1) ⇒ 2α + γ = – 5

⇒ γ = – 5 – 2α

(2) ⇒ α^{2} + αy + αy = \(\frac{-23}{4}\)

⇒ α^{2} + 2αγ = \(\frac{-23}{4}\)

⇒ α^{2} + α (- 5 – 2α) = \(\frac{-23}{4}\)

⇒ 4 (α^{2} – 10α – 4α^{2}) = – 23

⇒ 4 (- 3α^{2} – 10α) = – 23

⇒ 12α^{2} + 40α – 23 = 0

⇒ 12α^{2} + 46α – 6α – 23 = 0

⇒ 2α (6α + 23) – 1(6α + 23) = 0

⇒ (6α + 23) (2α – 1) = 0

∴ α = \(\frac{-23}{6}\) does not satisfy given equation.

∴ α = \(\frac{1}{2}\)

⇒ β = \(\frac{1}{2}\)

γ = – 5 – 2 . \(\frac{1}{2}\) = – 6

γ = – 6

∴ The roots of the given equation are \(\frac{1}{2}\), \(\frac{1}{2}\), – 6.

Question 12.

Solve x^{4} + 4x^{3} – 2x^{2} – 12x + 9 = 0, given that it has two pairs of equal roots. [March ’11, ’03]

Solution:

Given equation is x^{4} + 4x^{3} – 2x^{2} – 12x + 9 = 0

Given that it has two pairs of equal roots then let the roots of the given equation be

α + α + β + β = \(\frac{-4}{1}\)

⇒ 2 (α + β) = – 4

⇒ α + β = – 2

αα + αβ + βα + βα + αβ + β^{2} = \(\frac{-2}{1}\)

α^{2} + 4αβ + β^{2}

(α + β)^{2} + 2αβ = – 2

4 + 2αβ = – 2

2αβ = – 6

αβ = – 3

We know that,

(α – β)^{2} = (α + β)^{2} – 4αβ

= 4 – 4 (- 3)

= 4 + 12 = 16

α – β = 4

α + β = – 2

⇒ 2α = 2

⇒ α = 1

α + β = – 2

1 + β = – 2

β = – 3

∴ The roots of the given equation are 1, 1, – 3, – 3.

Question 13.

Find the roots of x^{4} – 16x^{3} + 86x^{2} – 176x + 105 = 0. [March ’08, ’02]

Solution:

Let f(x) = x^{4} – 16x^{3} + 86x^{2} – 176x + 105 = 0

If x = 1 then f(1)= 1 – 16 + 86 – 176 + 105 = 0

= 192 – 192 = 0

Hence, 1 is a root of f(x) = 0

If x = 3 then

f(3) = 81 – 16 (27) + 86(9) – 176(3) + 105

= 81 – 432+ 774 – 528 + 105

= 960 – 960 = 0

Hence 3 is a root of f(x) = 0.

By the method of synthetic division we have

∴ The quotient is x^{2} – 12x + 35 and remainder is zero.

∴ f(x) = x^{4} – 16x^{3} + 86x^{2} – 176x + 105

= (x – 1)(x – 3) (x^{2} – 12x + 35)

= (x – 1) (x – 3) (x – 5) (x – 7)

Hence the roots of the given equation are 1, 3, 5, 7.

Question 14.

Solve x^{4} – 10x^{3} + 26x^{2} – 10x + 1 = 0. [TS – Mar. ‘18; Mar. ‘12, ’10, Board Paper]

Solution:

Given equation is x – 10x^{3} + 26x^{2} – 10x + 1 = O

This is standard form of reciprocal equation.

The given equation is a reciprocal degree of class – 1 on dividing with x^{2}

⇒ x^{2} – 10x + 26 – \(\frac{10}{x}+\frac{1}{x^2}\) = 0

\(\left(x^2+\frac{1}{x^2}\right)-10\left(x+\frac{1}{x}\right)\) + 26 = 0

If x + \(\frac{1}{x}\) = a

⇒ x^{2} + \(\frac{1}{x^2}\) = a^{2} – 2

⇒ (a^{2} – 2) – 10a + 26 = 0

⇒ a^{2} – 10a + 24 = 0

⇒ a^{2} – 6a – 4a + 24 = 0

⇒ a (a – 6) – 4 (a – 6) = 0

If (a – 4) (a – 6) = 0

⇒ a = 4, a = 6

If a = 4

⇒ x + \(\frac{1}{x}\) = 4

⇒ x^{2} – 4x + 1 = 0

⇒ x = \(\frac{-(-4) \pm \sqrt{16-4(1)}}{2(1)}\)

= \(\frac{4 \pm \sqrt{12}}{2}\)

= 2 ± √3

If a = 6 x + \(\frac{1}{x}\) = 4

⇒ x^{2} – 6x + 1 = 0

x = \(\frac{-(-6) \pm \sqrt{36-4(1)(1)}}{2(1)}=\frac{6 \pm \sqrt{32}}{2}\)

= 3 ± 2√2

Hence ti roots of the given equation are 2 ± √3, 3 ± 2√2.

Question 15.

Solve the quation 6x^{4} – 35x^{3} + 62x^{2} – 35x + 6 = 0. [May ’13, ’10]

Solution:

Given equation is 6x^{4} – 35x^{3} + 62x^{2} – 35x + 6 = 0

This is a standard form of reciprocal equation.

The given equation is an even degree reciprocal equation of class – 1

On dividing both sides of given equation by x^{2}, we get

6x^{2} – 35x + 62 – \(\frac{35}{x}+\frac{6}{x^2}\) = 0

\(6\left(x^2+\frac{1}{x^2}\right)-35\left(x+\frac{1}{x}\right)\) + 62 = 0

Let x + \(\frac{1}{x}\) = a

⇒ \(\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\) = a^{2} – 2

⇒ 6(a^{2} – 2) – 35a + 62 = 0

⇒ 6a^{2} – 12 – 35a + 62 = 0

⇒ 6a^{2} – 35a + 50 = 0

⇒ 6a^{2} – 15a – 20a + 50 = 0

⇒ 3a (2a – 5) – 10 (2a – 5) = 0

⇒ (2a – 5)(3a – 10) = 0

⇒ 2a = 5; 3a = 10

a = \(\frac{5}{2}\) a = \(\frac{10}{3}\)

If a = \(\frac{5}{2}\)

⇒ x + \(\frac{1}{x}\) = \(\frac{5}{2}\)

⇒ 2x^{2} + 2 = 5x

⇒ 2x^{2} – 5x + 2 = 0

⇒ 2x^{2} – 4x – x + 2 = 0

⇒ 2x (x – 2) – 1 (x – 2) = 0

⇒ x = \(\frac{1}{2}\) (or) x = 2

If a = \(\frac{10}{3}\)

⇒ x + \(\frac{1}{x}\) = \(\frac{10}{3}\)

⇒ 3x^{2} + 3 = 10x

⇒ 3x^{2} – 10x + 3 = 0

⇒ 3x^{2} – 9x – x + 3 = 0

⇒ 3x (x – 3) – 1 (x – 3) = 0

x = \(\frac{1}{3}\) (or) x = 3

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

Question 16.

Solve 2x^{5} + x^{4} – 12x^{3} – 12x^{2} + x + 2 = 0. [TS – May ‘15: AP-Mar. ’17, ‘16; March ‘08, ‘07]

Solution:

Given equation is 2x^{5} + x^{4} – 12x^{3} – 12x^{2} + x + 2 = 0

Let f(x) = 2x^{5} + x^{4} – 12x^{3} – 12x^{2} + x + 2 = 0

If x = – 1

⇒ (- 1) = 2 (- 1) + 1 – 12 (- 1) – 12(1) + (- 1) + 2

= – 2 + 1 + 12 – 12 – 1 + 2 = 0

∴ (x + 1) is a factor of f(x).

On dividing f(x) by (x + 1) we get,

∴ 2x^{4} – x^{3} – 11x^{2} – x + 2 = 0

On dividing with x^{2} on both sides, we get

⇒ 2x^{2} – x – 11 – \(\frac{1}{x}+\frac{2}{x^2}\) = 0

⇒ \(2\left(x^2+\frac{1}{x^2}\right)-1\left(x+\frac{1}{x}\right)\) – 11 = 0

Let x + \(\frac{1}{x}\) = a

⇒ x^{2} + \(\frac{1}{x^2}\) + 2 = a^{2}

⇒ x^{2} + \(\frac{1}{x^2}\) = a^{2} – 2

⇒ 2(a^{2} – 2) – 1 (a) – 11 = 0

⇒ 2a^{2} – 4 – a – 11 = 0

⇒ 2a^{2} – a – 15 = 0

⇒ 2a^{2} – 6a + 5a – 15 = 0

⇒ 2a (a – 3) + 5 (a – 3) = 0

⇒ (a – 3) (2a + 5) = 0

a = 3; a = – \(\frac{5}{2}\)

If a = 3;

x + \(\frac{1}{x}\) = 3

x^{2} – 3x + 1 = 0

x = \(\frac{-(-3) \pm \sqrt{9-4}}{2(1)}=\frac{3 \pm \sqrt{5}}{2}\)

If a = – \(\frac{5}{2}\)

x + \(\frac{1}{x}\) = – \(\frac{5}{2}\)

⇒ 2x^{2} + 5x + 2 = 0

⇒ 2x^{2} + 4x + x + 2 = 0

⇒ 2x (x + 2) + 1 (x + 2) = 0

⇒ (2x + 1) (x + 2) = 0

∴ x = – \(\frac{1}{2}\), x = – 2

∴ The roots are – 1, – \(\frac{1}{2}\), – 2, \(\frac{3 \pm \sqrt{5}}{2}\).

Question 17.

Solve the equation x^{5} – 5x^{4} + 9x^{3} – 9x^{2} + 5x – 1 = 0. [AP – May 2016; March ‘13]

Solution:

Given equation is x^{5} – 5x^{4} + 9x^{3} – 9x^{2} + 5x – 1 = 0

Let f(x) = x^{5} – 5x^{4} + 9x^{3} – 9x^{2} + 5x – 1

If x = 1

f(1)= 1^{5} – 5(1)^{4} + 9(1)^{3} – 9(1)^{2} + 5(1) – 1

= 1 – 5 + 9 – 9 + 5 – 1 = 0

∴ (x – 1) is a factor of 1(x).

On dividing x^{5} – 5x^{3} + 9x^{3} – 9x^{2} + 5x – 1 by x – 1

∴ x^{4} – 4x^{3} + 5x^{2} – 4x + 1 = 0

This is a standard form of reciprocal equation.

Now, on dividing with x^{2} on both sides, we get

x^{2} – 4x + 5 – \(\frac{4}{x}+\frac{1}{x^2}\) = 0

\(\left(x^2+\frac{1}{x^2}\right)\) – 4 \(\left(x+\frac{1}{x}\right)\) + 5 = 0

Let x + \(\frac{1}{x}\) = a

⇒ [x + \(\frac{1}{x}\)]^{2} = a^{2}

⇒ x^{2} + \(\frac{1}{x^2}\) + 2 = a^{2}

⇒ x^{2} + \(\frac{1}{x^2}\) = a^{2} – 2

⇒ a^{2} – 2 – 4a + 5 = 0

⇒ a^{2} – 4a + 3 = 0

⇒ a^{2} – 3a – a + 3 = 0

⇒ a (a – 3) – 1 (a – 3) = 0

⇒ (a – 1) (a – 3) = 0

⇒ a = 1 (or) a = 3

If a = 1

⇒ x + \(\frac{1}{x}\) = 1

⇒ x^{2} – x + 1 = 0

⇒ x = \(\frac{-(-1) \pm \sqrt{1-4}}{2(1)}=\frac{1 \pm \mathrm{i} \sqrt{3}}{2}\)

If a = 3

⇒ x + \(\frac{1}{x}\) = 3

⇒ x^{2} – 3x + 1 = 0

⇒ x = \(\frac{-(-3) \pm \sqrt{9-4}}{2(1)}=\frac{3 \pm \sqrt{5}}{2}\)

∴ The roots are 1, \(\frac{1 \pm \mathrm{i} \sqrt{3}}{2}\), \(\frac{3 \pm \sqrt{5}}{2}\).

Question 18.

Solve the equation 6x^{6} – 25x^{5} + 31x^{4} – 31x^{2} + 25x – 6 = 0. [AP – Mar. ‘18. May 2015]

Solution:

Given equation is 6x^{6} – 25x^{5} + 31x^{4} – 31x^{2} + 25x – 6 = 0

The given equation is an even degree reciprocal equation of class – 2.

Let f(x) = 6x^{6} – 25x^{5} + 31x^{4} – 31x^{2} + 25x – 6 = 0

If x = 1 then f(1) = 6 – 25 + 31 – 31 + 25 – 6 = 0

∴ (x – 1) is a factor of f(x).

If x = – 1 then

f(- 1) = 6 + 25 + 31 – 31 – 25 – 6 = 0

∴ (x + 1) is a factor of f(x).

On dividing 1(x) by x + 1 and x – 1 we get,

∴ Quotient is 6x^{4} – 25x^{3} + 37x^{2} – 25x + 6

6x^{4} – 25x^{3} + 37x^{2} – 25x + 6 = 0

This is standard form of reciprocal equation.

This is an even degree reciprocal equation of class – 1.

On dividing both sides with x^{2}.

6x^{2} – 25x + 37 – \(\frac{25}{x}+\frac{6}{x^2}\) = 0

\(6\left(x^2+\frac{1}{x^2}\right)-25\left(x+\frac{1}{x}\right)\) + 37 = 0

Let x + \(\frac{1}{x}\) = a

\(x^2+\frac{1}{x^2}\) = a^{2} – 2

⇒ 6(a^{2} – 2) – 25a + 37 = 0

⇒ 6a^{2} – 25a + 25 = 0

⇒ 6a^{2} – 15a – 10a + 25 = 0

⇒ 3a(2a – 5) – 5 (2a – 5) = 0

⇒ (3a -5) (2a – 5) = 0

⇒ a = \(\frac{5}{2}\), a = \(\frac{5}{3}\)

If a = \(\frac{5{2}\)

⇒ x + \(\frac{1}{x}\) = a

⇒ x + \(\frac{1}{x}\) = \(\frac{5}{2}\)

⇒ 2x^{2} – 5x + 2 = 0

⇒ 2x^{2} – 4x – x + 2 = 0

⇒ 2x (x – 2) – 1 (x – 2) = 0

⇒ x = \(\frac{1}{2}\); x = 2

If a = \(\frac{5}{3}\)

⇒ x + \(\frac{1}{x}\) = \(\frac{5}{3}\)

⇒ 3x^{2} – 5x + 3 = 0

x = \(\frac{-(-5) \pm \sqrt{25-4(3)(3)}}{2(3)}\)

= \(\frac{5 \pm \sqrt{25-36}}{6}=\frac{5 \pm \mathrm{i} \sqrt{11}}{6}\)

∴ The roots are ± 1, 2, \(\frac{1}{2}\), \(\frac{5 \pm \mathrm{i} \sqrt{11}}{6}\).

Question 19.

Find the polynomial equation of degree 5 whose roots are the translates of the roots of x_{5} + 4x_{3} – x_{2} + 11 = 0 by – 3. [March ’06]

Solution:

Let f(x) = x^{5} + 4x^{3} – x^{2} + 11 = 0

∴ The required equation is f(x + 3) = 0

∴ f(x + 3) = A_{0}x^{5} + A_{1}x^{4} + A_{2}x^{3} + A_{3}x^{2} + A_{4}x + A_{5}

The coefficients A_{0}, A_{1}, A_{2}, A_{3}, A_{4}, A_{5} can be obtained as follows

∴ Required equation is f(x + 3) = 0

x_{5} + 15x_{4} + 94x^{3} + 305x_{2} + 507x + 353 = 0

Question 20.

Solve the equation x^{4} – 9x^{3} + 27x^{2} – 29x + 6 = 0 given that one root is 2 – √3. [May ’03]

Solution:

Given equation is x^{4} – 9x^{3} + 27x^{2} – 29x + 6 = 0

2 – √3 is a root of it then 2 + √3 is also root.

The quadratic factor of these two roots is (x – (2 + √3)) (x – (2 – √3))

= (x – (2 – √3)) (x – (2 + √3))

= ((x – 2) – √3) ((x – 2) + √3)

= (x – 2)^{2} – (√3)^{2}

= x^{2} – 4x + 4 – 3

= x^{2} – 4x + 1

On dividing x^{4} – 9x^{3} + 27x^{2} – 29x + 6 with x^{2} – 4x + 1

∴ Quotient is x^{2} – 5x + 6 = 0

∴ x^{2} – 5x + 6 = 0

⇒ x^{2} – 3x – 2x + 6 = 0

⇒ x(x – 3) – 2 (x – 3) = 0

⇒ (x – 2) (x – 3) = 0

⇒ x = 2 (or) x = 3

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

Question 21.

Show that x^{5} – 5x^{3} + 5x^{2} – 1 = 0, has three equal roots and find that root. [TS – Mar. 2017]

Solution:

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) and f(x).

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

Question 22.

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

Solution:

q^{3} = p^{3}r.

Question 23.

Solve 8x^{3} – 36x^{2} – 18x + 81 = 0, given that the roots are in AP. [March ‘04]

Solution:

\(\frac{-3}{2}, \frac{3}{2}, \frac{9}{2}\).

Question 24.

Solve 3x^{3} – 26x^{2} + 52x – 24 = 0, given that the roots are in A.P.

Solution:

\(\frac{2}{3}\), 2, 6

Question 25.

Solve the equation 6x^{3} – 11x^{2} + 6x – 1 = 0, the roots being in H.P.

Solution:

1, \(\frac{1}{2}\), \(\frac{1}{3}\)

Question 26.

Solve x^{3} – 7x^{2} + 36 = 0, gi0ven one root being twice the other.

Solution:

3, 6, – 2.

Question 27.

Solve x^{4} – 5x^{3} + 5x^{2} + 5x – 6 = 0, given that the product of two of its roots ¡s 3.

Solution:

2, – 1, 3, 1

Question 28.

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

Solution:

\(\frac{1}{3}\), \(\frac{1}{3}\), 1.

Question 29.

Solve x^{3} – 3x^{2} – 10x + 48 = 0.

Solution:

3, – 4, 4.

Question 30.

Solve the equation 4x^{3} – 13x^{2} – 13x + 4 = 0.

Solution:

– 1, \(\frac{1}{4}\), – 4

Question 31.

Find the polynomial equation whose roots are the translates of those of the equation. [TS – Mar. ’16]

x^{5} – 4x^{4} + 3x^{2} – 4x + 6 = Oby – 3.

Solution:

x^{5} + 11x^{4} + 42x^{3} – 57x^{2} – 13x – 60 = 0

Question 32.

Solve the equation x^{4} + 2x^{3} – 16x^{2} – 22x + 7 = 0 given that 2 – √3 is one of its roots.

Solution:

2 ± √3, – 3 ± √2.

### Some More Maths 2A Theory of Equations Important Questioons

Question 1.

Form the monic polynomial equation of degree 3 whose roots are 2, 3 and 6. [March ‘02]

Solution:

Let α = 2, β = 3, γ = 6

The monk equation having roots α, β, γ is

(x – α) (x – β) (x – γ) = 0

⇒ (x – 2) (x – 3) (x – 6) = 0

⇒ (x – 2) (x^{2} – 9x + 18) =0

⇒ x^{3} – 9x^{2} + 18x – 2x^{2} + 18x – 36 = 0

⇒ x^{3} – 11x^{2} + 36x – 36 = 0.

Question 2.

Form the monic polynomial equation of degree 4 whose roots are 4 + √3, 4 – √3, 2 + i and 2 – i.

Solution:

Let α = 4 + √3, β = 4 – √3, γ = 2 + i, δ = 2 – i

The equation whose roots are α, β, γ, δ is

(x – α) (x – β) (x – γ) (x – δ) = 0

(x – (4 + √3)) (x – (4 – √3))

(x – (2 + i)) (x – (2 – i)) = 0

((x – 4) – √3) ((x – 4) + √3)

((x – 2) – i) ((x – 2) + i) = 0

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

(x^{2} + 16 – 8x – 3) (x^{2} + 4 – 4x + 1) = 0

(x^{2} – 8x + 13) (x^{2} – 4x + 5) = 0

x^{4} – 12x^{3} + 50x^{2} – 92x + 65 = 0

Question 3.

Find s_{1}, s_{2}, s_{3} and s_{4} for the equation 8×4 – 2×3 – 27×2 – 6x + 9=0.

Solution:

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

Comparing this equation with ax^{4} + bx^{3} + cx^{2} + dx + e = 0

a = 8; b = – 2; c = – 27; d = – 6; e = 0

Now,

s_{1} = \(\frac{-b}{a}=\frac{-(-2)}{8}=\frac{1}{4}\)

s_{2} = \(\frac{c}{a}=\frac{-27}{8}\)

s_{3} = \(\frac{-\mathrm{d}}{\mathrm{a}}=\frac{-(-6)}{8}=\frac{3}{4}\)

s_{4} = \(\frac{\mathrm{e}}{\mathrm{a}}=\frac{9}{8}\)

Question 4.

Find the algebraic equation whose roots are 3 times the roots of x^{3} + 2x^{2} – 4x + 1 = 0.

Solution:

Let f(x) = x^{3} + 2x^{2} – 4x + 1 = 0

∴ Required equation is f(\(\frac{x}{3}\)) = 0

\(\) = 0

⇒x3 + 6×2-36x + 27=0.

Question 5.

Find an algebraic equation of degree 4 whose roots are 3 times the roots of the equation 6x^{4} – 7x^{3} + 8x^{2} – 7x + 2 = 0.

Solution:

Let f(x) = 6x^{4} – 7x^{3} + 8x^{2} – 7x + 2 = 0

The required equation is f(\(\frac{x}{3}\)) = 0

\(\frac{6 x^4}{81}-\frac{7 x^3}{27}+\frac{8 x^2}{9}-\frac{7 x}{3}+2\) = 0

6x^{4} – 21x^{3} + 72x^{2} – 189x + 162 = 0

Question 6.

Find the transfonned equation whose roots are the negatives of the roots of x^{4} + 5x^{3} + 11x + 3 = 0.

Solution:

Let f(x) = x^{4} + 5x^{3} + 11x + 3 = 0

Required equation ¡s f(- x) = 0

⇒ (- x)^{4} + 5(- x)^{3} + 11(- x) + 3 = 0

⇒ x^{4} – 5x^{3} – 11x + 3 = 0

Question 7.

Find the polynomial equation of degree 4 whose roots are the negatives of the roots of x^{4} – 6x^{3} + 7x^{2} – 2x + 1 = 0.

Solution:

Let f(x) = x^{4} – 6x^{3} + 7x^{2} – 2x + 1 = 0

The required equation is f(- x) = 0

⇒ (- x)^{4} – 6(- x)^{3} + 7(- x)^{2} – 2(- x) + 1 = 0

⇒ x^{4} + 6x^{3} + 7x^{2} + 2x + 1 = 0.

Question 8.

Find the polynomial equation whose roots are the reciprocals of the roots of x^{5} + 11x^{4} + x^{3} + 4x^{2} – 13x + 6 = 0

Solution:

Let f(x) = x^{5} + 11x^{4} + x^{3} + 4x^{2} – 13x + 6 = 0

The required equation is f(\(\frac{1}{x}\)) = o

⇒ \(\frac{1}{x^5}+\frac{11}{x^4}+\frac{1}{x^3}+\frac{4}{x^2}-\frac{13}{x}+6\) + 6 = 0

⇒ 1 + 11x + x^{2} + 4x^{3} – 13x^{4} + 6x^{5} = 0

⇒ 6x^{5} – 13x^{4} + 4x^{3} + x^{2} + 11x + 1 = 0

Question 9.

Find the polynomial equation whose roots are the reciprocals of the roots of the equation x^{4} + 3x^{3} – 6x^{2} + 2x – 4 = 0.

Solution:

Let f(x) = x^{4} + 3x^{3} – 6x^{2} + 2x – 4 = 0

The required equation is f(\(\frac{1}{x}\))) = 0

⇒ \(\frac{1}{x^4}+\frac{3}{x^3}-\frac{6}{x^2}+\frac{2}{x}-4\) = 0

⇒ 4x^{4} – 2x^{3} + 6x^{2} – 3x – 1 = 0

Question 10.

Find the polynomial equation whose roots are the squares of the roots of x^{4} + x^{3} + 2x^{2} + x + 1 = 0.

Solution:

Let f(x) = x^{4} + x^{3} + 2x^{2} + x + 1 = 0

The required equation is f(√x) = 0

(√x)^{4} + (√x)^{3} + 2(√x)^{2} + √x + 1 = 0

x^{2} + x√x + 2x + √x + 1 = 0

x^{2} + 2x + 1 = – x(√x) – √x

= – √x (x + 1)

⇒ squaring on both sides

(x^{2} + 2x + 1)^{2} = (- √x (x + 1))^{2}

⇒ x^{4} + 4x^{2} + 1 + 4x^{3} + 4x + 2x^{2} = x(x^{2} + 1 + 2x) = x^{3} + x + 2x^{2}

x^{4} + 3x^{3} + 4x^{2} + 3x + 1 = 0.

Question 11.

Find the polynomial equation whose roots are the squares of the roots of x^{3} – x^{2} + 8x – 6 = 0.

Solution:

Let f(x) = x^{3} – x^{2} + 8x – 6 = 0

∴ The required equation is f(√x) = o

(√x)^{3} – (√x)^{2} + 8√x + 6 = 0

⇒ x√x – x + 8√x – 6 = 0

⇒ x + 6 = √x (x + 8)

⇒ squaring on both sides

⇒x^{2} + 36 + 12x = x(x^{2} + 64 + 16x)

⇒ x^{3} + 64x + 16x^{2}

⇒ x^{3} + 15x^{2} + 52x – 36 = 0.

Question 12.

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

Solution:

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

Since the roots are in G.P. then they must be the of the form \(\frac{a}{r}\), a, ar.

Now s_{3} = \(\frac{a}{r}\) . a . ar

= \(\frac{-(-r)}{1}\)

a^{3} = r

a = r^{1/3}

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

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

⇒ r – pr^{2/3} + qr^{1/3} – r = 0

⇒ pr^{2/3} = qr^{1/3}

Cubing on both sides.

⇒ p^{3}r^{2} = q^{3}r

⇒ p^{3}r = q^{3}

Question 13.

Solve 8x^{3} – 36x^{2} – 18x + 81 = 0, given that the roots are in AP.

Solution;

Given equation is 8x^{3} – 36x^{2} – 18x + 81 = 0

Since, the roots are in A.P, they must be of the form a – d, a, a + d.

Now, a – d + a + a + d = \(\frac{-(-36)}{8}\)

⇒ 3a = \(\frac{9}{2}\)

⇒ a = \(\frac{3}{2}\)

(a – d) a (a + d) = \(\frac{-81}{8}\)

(a^{2} – d^{2}) a = \(\frac{-81}{8}\)

\(\frac{3}{2}\left(\frac{9}{4}-\mathrm{d}^2\right)=\frac{-81}{8}\)

\(\frac{9}{4}-d^2=\frac{-27}{4}\)

d^{2} = 36

⇒ d = ± 3.

Case – I:

a = \(\frac{3}{2}\), d = 3

The roots of the given equation are a – d, a, a + d

\(\frac{-3}{2}, \frac{3}{2}, \frac{9}{2}\)

Case – II: a = \(\frac{3}{2}\), d = – 3

The roots of the given equation are a – d, a, a + d

\(\frac{9}{2}, \frac{3}{2}, \frac{-3}{2}\)

Question 14.

Solve 3x^{3} – 26x^{2} + 52x – 24 = 0 the given, the roots are in Geometric Progression.

Solution:

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

Since the roots are in G.P they must be of the form \(\frac{a}{r}\), a, ar.

Now, \(\frac{a}{r}\) + a + ar = \(\frac{-(-26)}{3}\)

a(\(\frac{1}{r}\) + 1 + r) = \(\frac{26}{3}\) …………..(1)

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

a^{3} = 8

a = 2

From (1)

⇒ \(2\left(\frac{1+r+r^2}{r}\right)=\frac{26}{3}\)

⇒ 3(r^{2} + r + 1) = 13r

⇒ 3r^{2} – 10r + 3 = 0

⇒ 3r^{2} – 9r – r + 3 = 0

⇒ 3r (r – 3) – 1(r – 3) = 0

⇒ (r – 3) (3r – 1) = 0

r = 3 (or) r = \(\frac{1}{3}\)

Case – 1:

a = 2, r = 3

The roots of the given equation are \(\frac{a}{r}\), a, ar = \(\frac{2}{3}\), 2, 6

Case – II:

a = 2, r = \(\frac{1}{3}\)

The roots of the given equation are \(\frac{a}{r}\), a, ar = 6, 2, \(\frac{2}{3}\).

Question 15.

Solve the equation 6x^{3} – 11x^{2} + 6x – 1 = 0, the roots being in 1-LP.

Solution:

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

The roots of 6x^{3} – 11x^{2} + 6x – 1 = 0 are in H.P

The roots of \(6 \cdot \frac{1}{x^3}-11 \cdot \frac{1}{x^2}+\frac{6}{x}-1\) = 0 are in A.P.

6 – 11x + 6x^{2} – x^{3} = 0

x^{3} – 6x^{2} + 11x – 6 = 0

Let the roots be a – d, a, a + d

a – d + a + a + d = \(\frac{-(-6)}{1}\)

3a = 6

a = 2

(a – d) a (a + d) = \(\frac{-(-6)}{1}\) = 6

⇒ 2(a^{2} – d^{2}) = 6

⇒ 4 – d^{2} = 3

⇒ d^{2} = 1

⇒ d = 1

The roots of x^{3} – 6x^{2} + 11x – 6 = 0 are a – d, a, a + d = 1, 2, 3.

∴ The roots of 6x^{3} – 11x^{2} + 6x – 1 = 0 are 1, \(\frac{1}{2}\), \(\frac{1}{3}\).

Question 16.

Solve x^{3} – 7x^{2} + 36 = 0, given one root being twice the other.

Solution;

Given equation is x’ – 7×2 + 36 = 0

Let α, β, γ be the roots of the given equation.

Now, α + β + γ = \(\frac{-(-7)}{1}\)

⇒ α + β + γ = 7 …………..(1)

αβ + βγ + γα = \(\frac{0}{1}\) = 0

⇒ αβ + βγ + γα = 0 ………….(2)

αβγ = \(\frac{-36}{1}\) = – 36

⇒ αβγ = – 36 ……………(3)

Given that one root being twke the other then β = 2α

(1) ⇒ α + 2α + γ = 7

⇒ 3α + γ = 7

⇒ α = 7 – 3α

(2) ⇒ α . 2α + 2αγ + αγ = 0

⇒ 2α + 3αγ = o

⇒ 2α^{2} + 3α (7 – 3α) = 0

⇒ 2α^{2} + 21α – 9α^{2} = 0

⇒ 7α^{2} – 21α = 0

⇒ α^{2} – 3α =0

⇒ α (α – 3) = 0

⇒ α = 0 (or) α = 3

α = 0 does not satisfy the given equation.

∴ α = 3

If α =3

⇒ β = 6

γ = 7 – 3α

= 7 – 3 . 3

= 7 – 9

⇒ γ = – 2

∴ The roots of the given equation are 3, 6, – 2.

Question 17.

Solve x^{4} – 5x^{3} + 5x^{2} + 5x – 6 = 0, given that the product of two of its roots is 3.

Solution:

Let α, β, γ, δ are the roots of given equation.

⇒ αβγδ = – 6

Given that the product of two roots is 3

αβ = 3

⇒ 3γδ = – 6

⇒ γδ = – 2

Let α + β = p, γ + δ = q

The quadratic equation whose roots are α, β is

x^{2} – x(α + β) + αβ = 0

⇒ x^{2} – px + 3 = 0 …………….(1)

The quadratic equation ol roots γ, δ is

x^{2} – x(γ + δ) + γδ = 0

⇒ x^{2} – qx – 2 = 0 ………………(2)

∴ x^{4} – 5x^{3} + 5x^{2} + 5x – 6 = (x^{2} – px + 3) (x^{2} – qx – 2)

= x^{4} – qx^{3} – 2x^{2} – px^{3} + pqx^{2} + 2px + 3x^{2} – 3qx – 6

= x^{4} + x^{ 3} (- p – q) + x^{2} (1 + pq) + x (2p – 3q) – 6

x^{3}coeff. ⇒ – p – q = – 5

⇒ p + q = 5 ……………..(3)

xcoeff. ⇒ 2p – 3q = 5 …………..(4)

Solving (3) and (4)

p = 4, q = 1

Sub. p = 4 in (1)

⇒ x^{2} -4x + 3 = 0

⇒ x^{2} – 3x – x + 3 = 0

⇒ x (x – 3) – 1 (x – 3) = 0

⇒ (x – 1) (x – 3) = 0

⇒ x – 1 = 0 (or) x – 3 = 0

⇒ x = 1 (or) x = 3

Sub.q = 1 in (2)

⇒ x^{2} – x – 2 = 0

⇒ x^{2} – 2x + x – 2 = 0

⇒ x (x – 2) + 1 (x – 2) = 0

⇒ (x + 1) (x – 2) = 0

⇒ x = 2 (or) x = – 1

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

Question 18.

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

Solution:

Given equation is 9x^{3} – 15x^{2} + 7x – 1 = 0

Let α, β, γ are the roots of 9x – 15x^{2} + 7x – 1 = 0

Now, α + β + γ = \(\frac{-(-15)}{9}=\frac{5}{3}\) ⇒ 1

⇒ αβ + βγ + γα = \(\frac{7}{9}\) …………….(2)

αβγ = \(\frac{-(-1)}{9}=\frac{1}{9}\) …………(3)

Given α = β; (1)

⇒ 2α + γ = \(\frac{5}{3}\) ……………(4)

(2) ⇒ α^{2} + αγ + γα = \(\frac{7}{9}\)

α^{2} + 2αγ = \(\frac{7}{9}\)

γ = \(\frac{7}{18 \alpha}-\frac{\alpha}{2}\) ………….(5)

From (4) 2α + \(\frac{7}{18 \alpha}-\frac{\alpha}{2}=\frac{5}{3}\)

⇒ \(\frac{36 \alpha^2+7-9 \alpha^2}{18 \alpha}=\frac{5}{3}\)

⇒ 27α^{2} – 30α + 7 = 0

⇒ 27α^{2} – 21α – 9α + 7 = 0

⇒ 3α (9α – 7) – 1 (9α – 7) = 0

⇒ (3α – 1) (9α – 7) = 0

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

α = does not satisfy the given equation.

∴ α = \(\frac{1}{3}\), β = \(\frac{1}{3}\)

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

γ = \(\frac{5}{3}-\frac{2}{3}=\frac{3}{3}\) = 1

γ = 1

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

Question 19.

Solve x^{3} – 3x^{2} – 16x + 48 = 0.

Solution:

Let f(x) = x^{3} – 3x^{2} – 16x + 48 = 0

If x = 3

⇒ f(3) = 27 – 27 – 48 + 48 = 0

Hence 3 is a root of f(x) = 0.

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

∴ The quotient is x^{2} – 16 and the remainder is zero.

∴ f(x) = x^{3} – 3x^{2} – 16x + 48

= (x – 3) (x^{2} – 16)

= (x – 3) (x + 4) (x – 4)

∴ The roots are 3, – 4, 4.

Question 20.

Solve the equation 4x^{3} – 13x^{2} – 13x + 4 = 0.

Solution:

Given equation is 4x^{3} – 13x^{2} – 13x 4 = 0.

f(x) = 4x^{3} – 13x^{2} – 13x + 4

If x = – 1

⇒ f(- 1) = 4(- 1)^{3} – 13(- 1)^{2} – 13(- 1) + 4

= – 4 – 13 + 13 + 4 = 0

∴ (x + 1) is a factor of f(x).

Now, on dividing 4x^{3} – 13x^{2} – 13x + 4 = 0 by x + 1.

Quotient is 4x^{2} – 17x + 4

4x^{2} – 17x + 4 = 0

4x^{2} – 16x – x + 4 = 0

=4x (x – 4) – 1 (x – 4) = 0

(4x – 1) (x – 4) = 0

x = \(\frac{1}{4}\), x = 4

∴ 4x^{3} – 13x^{2} – 13x + 4 = (x + 1) (x – 4) (4x – 1)

∴ The roots of the given equation are – 1, 4, 1/4.

Question 21.

Find the polynomial equation whose roots are the translates of those of the equatior x^{5} – 4x^{4} + 3x^{2} – 4x + 6 = 0 by – 3.

Solution:

Let f(x) = x^{5} – 4x^{4} + 3x^{2} – 4x + 6

∴ The required equation is f(x + 3) =0

∴ f(x+3) = A_{0}x^{5} + A_{1}x^{4} + A_{2}x^{3} + A_{3}x^{2} + A_{4}x + A_{5}

The coefficients A_{0}, A_{1}, A_{2}, A_{3}, A_{4}, A_{5} can be obtained as follows.

∴ Required equation ¡s f(x + 3) = 0

x^{5} + 11x^{4} + 42x^{3} + 57x^{2} – 13x – 60 = 0.

Question 22.

Solve the equation x^{4} + 2x^{3} – 16x^{2} – 22x + 7 = 0 given that 2 – √3 is one of its roots.

Solution:

Given equation is x^{4} + 2x^{3} – 16x^{2} – 22x + 7 = 0 ……………..(1)

Let f(x) = x^{4} + 2x^{3} – 16x^{2} – 22x + 7

Given 2 – √3 is a root of (1).

⇒ 2 + √3 is also a root of (1).

(∵ coefficients of (1) are rational)

∴ (x – (2 – √3)) (x – (2 + √3)) is a factor of f(x).

(x^{2} – 4x + 1)is a factor of f(x).

We divide f(x) by x^{2} – 4x 1.

By synthetic division,

∴ f(x)=(x^{2} – 4x +1 ) (x^{2} + 6x + 7)

∴ Equation (1)

z(x^{2} – 4x + 1) (x^{2} + 6x + 7) = 0

=x^{2} – 4x + 1 = 0 or x^{2} + 6x + 7 = 0

x = 2 ± √3 or x = – 3 ± √2

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

Question 23.

If α, β, γ are the roots of x^{3} + px^{2} + qx + r = 0, then form the monic cubic equation whose roots are α(β + γ), β(γ + α), γ(α + β).

Solution:

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

Since α, β, γ are the roots of the equation.

x^{3} + px^{2} + qx + r = 0 then

s_{1} = α + β + γ = \(\frac{-p}{1}\) = – p;

s_{2} = αβ + βγ + γα = \(\frac{q}{1}\) = q;

s_{3} = αβγ = \(\frac{-r}{1}\) = – r

Let s_{1} = Σ α(β + γ)

= Σ (αβ + αγ) = Σ αβ + Σ αγ = Σ αβ + Σ αβ = 2Σ αβ = 2q

s_{2} = Σ α(β + γ) . β(γ + α)

= Σ αβ (- p – α) (- p – β)

= Σ αβ (p + α) (p + β)

= Σ αp (p^{2} + p(α + β) + αβ)

= Σ αp [p^{2} + p(- p – γ) + ap]

= Σ αβ [p^{2} – p^{2} – pγ + αβ]

= – pΣ αβγ + Σ αβ

= – p(3αβγ) + (αβ + βγ + γα)^{2} – 2αβγ(α + β + γ)

= – 3p (- r) + q^{2} – 2 (- r) (- p)

= 3pr + q^{2} – 2pr

= q^{2} + pr

s_{3} = α (β + γ) β (γ + α) γ (α + β)

= αβγ(- p – α) (- p – β) (- p – γ)

= – αβγ [(p + α) (p + β) (p + γ)]

= – αβγ [(p^{2} + pα + pβ + αβ) (p + γ)]

= – αβγ [p^{3} + p^{2}α + p^{2}β + pαβ + p^{2}γ + pαγ + pβγ + αβγ]

= – αβγ [p^{3} + p^{2} (α + β + γ) + p(αβ + βγ + γα) + αβγ]

= – (- γ)[p^{3} + p^{2} (- p) + p(q) + (- r)]

= γ [p^{3} – p^{3} + pq – r]

= pqr – r^{2}

∴ The monic cubic equation whose roots are α (β + γ), β(γ + α), γ(α + β) is

x^{3} – s_{1}x^{2} + s_{2}x – s_{3} = 0

x^{3} – 2qx^{2} + (q^{2} + pr) x – r (pq – r) = 0.

Question 24.

If α, β, γ are the roots of x^{3} – 6x^{2} + 11x – 6 = 0, then find the equation whose roots are α^{2} + β^{2}, β^{2} + γ^{2}, γ^{2} + α^{2}.

Solution:

Let f(x) = x^{3} – 6x^{2} + 11x – 6

If x = 1 then f(1) = 1 – 6 + 11 – 6 = 12 – 12 = 0

(x – 1) is the factor of f(x)

x^{2} – 5x + 6 = 0

⇒ x^{2} – 3x – 2x + 6 = 0

⇒ x (x – 3) – 2 (x – 3) = 0

⇒ (x – 2) (x – 3) = 0

⇒ x – 2 = 0; x – 3 = 0

Therefore, x^{3} – 6x^{2} + 11x – 6 = 0

⇒ (x – 1) (x – 2) (x – 3) = 0

⇒ x = 1, 2, 3

Since α, β, γ are the roots of x^{3} – 11x^{2} + 11x – 6 = 0

Then, α = 1; β = 2; γ = 3.

Now, α^{2} + β^{2} = 1 + 4 = 5;

β^{2} + γ^{2} = 4 + 9 = 13;

γ^{2} + α^{2} = 9 + 1 = 10

The equation having roots 5, 13, 10 is (x – 5) (x – 13) (x – 10) = 0

= (x^{2} – 18x + 65) (x – 10) = 0

x^{3} – 18x^{2} + 65x – 10x^{2} + 180x – 650 = 0

x^{3} – 28x^{2} + 245x – 650 = 0

Question 25.

If α, β, γ are the roots of x^{3} – 7x + 6 = 0, then find the equation whose roots are (α – β)^{2}, (β + γ)^{2}, (γ + α)^{2}.

Solution:

Let f(x) = x^{3} – 7x + 6

If x = 1 then f(1) = 1 – 7(1) + 6 – 7 – 7 = 0

∴ (x – 1) is a factor of f(x)

x^{2} + x – 6 = 0

⇒ x^{2} + 3x – 2x – 6 = 0

x (x + 3) – 2 (x + 3) = 0

(x + 3) (x – 2) = 0

∴ x^{3} – 7x + 6 = 0

⇒ (x + 3) (x – 1) (x – 2) = 0

⇒ x = – 3, 1, 2

Since α, β, γ are the roots of x^{3} – 7x + 6 = 0

then, α = – 3; β = 1; γ = 2.

Now, (α – β)^{2} = (- 3 – 1)^{2} = 16;

(β – γ)^{2} = (1 – 2)^{2} = 1;

(γ – α)^{2} = (4 + 9)^{2} = 25

The equation having roots 16, 1, 25 is (x – 16) (x – 1) (x – 25) = 0

(x^{2} – 1 7x + 16) (x^{2} – 25) = 0

x^{3} – 17x^{2} + 16x – 25x^{2} + 465x – 400 = 0

x^{3} – 42x^{2} + 441x – 400 = 0.

Question 26.

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

Solution:

Let the roots be 3α, 2α, β

3α + 2α + β = \(\frac{-(-9)}{1}\) = 9

⇒ 5α + β = 9

⇒ β = 9 – 5α ……………..(1)

3α . 2α + 2αβ + 3αβ = \(\frac{14}{1}\)

⇒ 6α^{2} + 5αβ = 14

6α^{2} + 5α – 14 = 0

⇒ 6α^{2} + 5α (9 – 5α)-14 =0

6α^{2} + 45α – 25α – 14 = 0

⇒ 19α^{2} + 45α – 14 = 0

⇒ 19α^{2} – 45α + 14 = 0

⇒ 19α^{2} – 38α – 7α + 14 = 0

⇒ 19α(α – 2) – 7(α – 2) = 0

(α – 2) (19α – 7) = 0

α = 2 (or)

α = \(\frac{7}{19}\)

⇒ α = \(\frac{14}{19}\)

⇒ 2α = does not satisfy the given equation.

∴ α = 2

3α = 6

2α = 4

(1) ⇒ β = 9 – 5(2)

(2) ⇒ 9 – 10 = – 1

β = – 1

∴ The roots of the given equation are 6, 4,- 1.

Question 27.

Solve the equation x^{4} – 6x^{3} + 13x^{2} – 24x + 36 = 0 given that they have multiple roots.

Solution:

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

if x = 3 then 81 – 6(27) + 13(9) – 24 (3) + 36

= 81 – 162 + 117 – 72 + 36

= – 234 + 234 = 0

∴ x = 3 is a root of f(x) = 0

Let g(x) = x^{3} – 3x^{2} + 4x – 12 = 0

If x = 3 then g(3) = 3^{3} – 3(3^{2}) + 4(3) – 12

= 27 – 27 + 12 – 12 = 0

Let, h(x) = x^{2} + 4 = 0

⇒ x^{2} = – 4

x = ± √4 = ± 2

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

Multiple root is 3.

Question 28.

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(\(\left(\frac{3}{2}\right)\)) = 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 \(-\frac{1}{2}, \frac{3}{2}, \frac{3}{2}\).

Question 29.

Solve 6x^{4} – 13x^{3} – 35x^{2} – x + 3 = 0, given that one of its roots is 2 + √3.

Solution:

Given equation is 6x^{4} – 13x^{3} – 35x^{2} – x + 3 = 0

Since 2 + √3 is a root of it, 2 – √3 is also a root.

The quadratic factor of these two roots is (x – (2 + √3)) (x – (2 – √3))

= ((x – 2) – √3) ((x – 2) + √3)

= (x – 2)^{2} – 3

= x^{2} + 4 – 4x – 3

= x^{2} – 4x + 1

On dividing, 6x^{4} – 13x^{3} – 35x^{2} – x + 3 by x^{2} – 4x + 1.

∴ Quotent is 6x^{2} + 11x + 3

6x^{2} + 11x + 3 = 0

6x^{2} + 9x + 2x + 3 = 0

3x (2x + 3) + 1 (2x + 3) = 0

= (3x + 1) (2x + 3) = 0

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

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

Question 30.

Solve the equation x^{4} + 2x^{3} – 5x^{2} + 6x + 2 = 0 given that 1 + i is one of its roots.

Solution:

Given equation is x^{4} + 2x^{3} – 5x^{2} + 6x + 2 = 0

Since 1 + i is a root of the given equation

then 1 – i is also a root of it.

The quadratic factor to these two roots is

(x – (1 + i)) ((x – (1 – i)) = 0

((x – 1) – i) ((x – 1) + i) = (x – 1) – i

x^{2} + 1 – 2x + 1 = x^{4} – 2x + 2

On dividing x^{4} + 2x^{3} – 5x^{2} 6x + 2 by x^{2} – 2x + 2

Quotient is x^{2} + 4x + 1

x^{2} + 4x + 1 = 0

x = \(\frac{-4 \pm \sqrt{1-4}}{2(1)}\) = – 2 ± √3

∴ The roots of the given equation are 1 + i, 1 – i, – 2 ± √3.

Question 31.

Solve x^{4} – 4x^{2} + 8x + 35 = 0, given that 2 + √3 is a root. [AP – Mar. 2015]

Solution:

Given equation is x^{4} – 4x^{2} + 8x + 35 = 0

Since 2 + i√3 is a root of it,

2 – i√3 is also a root.

The quadratic factor to these two roots is (x – (2 + i√3)) (x – (2 – i√3))

= ((x – 2) – i√3) ((x – 2) + i√3)

= (x – 2)^{2} (i√3)^{2}

= x^{2} + 4 – 4x + 3

= x^{2} – 4x + 1

On dividing x^{4} – 4x^{2} + 8x + 35 with x^{2} – 4x + 7 we get,

∴ Quotient is x^{2} + 4x + 5

x^{2} + 4x + 5 = 0

x = \(\frac{-4 \pm \sqrt{16-20}}{2(1)}=\frac{-4 \pm \sqrt{-4}}{2}\) = – 2

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

Question 32.

Solve the equation x^{4} – 6x^{3} + 11x^{2} – 10x + 2 = 0, given that 2 + √3 is a root of the equation.

Solution:

Given equation is x^{4} – 6x^{3} + 11x^{2} – 10x + 2 = 0

Since 2 + √3 is a root of it, 2 – √3 is also a root.

The quadratic factor to these roots is (x – (2 + √3)) (x – (2 – √3))

= ((x – 2) – √3) ((x – 2) + √3)

= (x – 2)^{2} – 3

= x^{2} + 4 – 4x – 3

= x^{2} – 4x + 1

∴ Quotient is x^{2} – 2x + 2

∴ x^{2} – 2x + 2 = 0

x = \(\frac{-(-2) \pm \sqrt{4-4(2)}}{2(1)}\)

= \(\frac{2 \pm 2 \mathrm{i}}{2}\) = 1 ± i.

∴ Roots of given equation are 2 ± √3, 1 ± i.

Question 33.

Find the polynomial equation whose roots are the translates of those of the equation x^{4} – 5x^{3} + 7x^{2} – 17x + 11 = 0 by – 2. [TS – May 2016]

Solution:

Let f(x) = x^{4} – 5x^{3} + 7x^{2} – 17x + 11 = 0

∴ The required equation is f(x + 2) = 0

Now, f(x + 2) = A_{0}x^{4} + A_{1}x^{3} + A_{2}x^{2} + A_{3}x + A_{4}

Then, the coefficients A_{0}, A_{1}, A_{2}, A_{3}, A_{4} can

be obtained as follows

∴ Required equation is f(x + 2) = 0

x^{4} + 3x^{3} + x^{2} – 17x – 19 = 0.

Question 34.

Find the polynomial equation whose roots are the translates of the roots of the equation x^{4} – x^{3} – 10x^{2} + 4x + 24 = 0 by 2.

Solution:

Let f(x) = x^{4} – x^{3} – 10x^{2} + 4x + 24 = 0

∴ Required equation is f(x -2) = 0

Let f(x – 2) = A_{0}x^{4} + A_{1}x^{3} + A_{2}x^{2} + A_{3}x + A_{4}.

The coefficients A_{0}, A_{1}, A_{2}, A_{3}, A_{4} can be obtained as follows

∴ Required equation is f(x – 2) = 0

x^{4} – 9x^{3} + 40x^{2} – 80x + 80 = 0.