Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Theory of Equations Important Questions which are most likely to be asked in the exam.

## Intermediate 2nd Year Maths 2A Theory of Equations Important Questions

Question 1.

Form polynomial equation of the lowest degree, with roots 1, -1, 3 (May ’06)

Hint : Equation having roots, α, β, γ is [(x – α)(x – β)(x – γ) = 0

Solution:

Required equation is

(x – 1) (x + 1) (x – 3) = 0

⇒ (x^{2} – 1) (x – 3) = 0

⇒ -3x^{2} – x + 3 = 0

Question 2.

If 1, 1 α are the roots of

x^{3} – 6x^{2} + 9x – 4 = 0, then find α. (May ’11)

Solution:

1, 1, α are roots of x^{3} – 6x^{2} + 9x – 4 = 0

Sum = 1 + 1 + α = 6

α = 6 – 2 = 4

Question 3.

If -1, 2 and α are the roots of 2x^{3} + x^{2} – 7x – 6 = 0, then find α (Mar 14, 13)

Solution:

-1, 2, α are roots of 2x^{3} + x^{2} – 7x – 6 = 0

sum = -1 + 2 + α = –\(\frac{1}{2}\)

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

Question 4.

If 1, -2 and 3 are roots of x^{3} – 2x^{2} + ax + 6 = 0, then find α. (Mar. ’04,)

Solution:

1, -2 and 3 are roots of x^{3} + x^{2} + ax – 6 = 0

x^{3} – 2x + ax + 6 = 0

⇒ 1(-2) + (-2) 3 + 3. 1 = a

i.e., a = -2 – 6 + 3 = -5

Question 5.

If the product of the roots of 4x^{3} + 16x^{2} – 9x – a = 0 is 9, then find a. (TS Mar. ’17, ’16)

Solution:

α, β, γ are the roots of

4x^{3} + 16x^{2} – 9x – a = 0

αβγ = \(\frac{a}{4}\) = 9 ⇒ a = 36

Question 6.

Find the transformed equation whose roots are the negative of the roots of

x^{4} + 5x^{3} + 11x + 3 = 0

Solution:

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

We want an equation whose roots are

-α_{1}, α_{2}, α_{3}, α_{4},

Required equation f(-x) = 0

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

⇒ x^{4} – 5x^{3} – 11x + 3 = 0

Question 7.

Form the polynomial equation of degree 3 whose roots are 2, 3 and, 6.

Solution:

The required polynomial equation is,

(x – 2) (x – 3) (x – 6) = 0

⇒ x^{3} – 11x^{2} + 36x – 36 = 0

Question 8.

Let α, β, γ be the roots of Σα^{3}

Solution:

Σα^{3} = α^{3} + β^{3} + γ^{3}

= (α + β + γ)

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

= (-p)(p^{2} – 2q – q) – 3r

= -p(p^{2} – 3q) – 3r

∴ Σα^{3} = -p^{3} + 3pq – 3r = 3pq – p^{3} – 3r

Question 9.

If α, β and 1 are the roots of x^{3} – 2x^{2} – 5x + 6 = 0, then find α and β (AP Mår ’16, ’08)

Solution:

α, β and 1 are the roots of

x^{3} – 2x^{2} – 5x + 6 = 0

Sum = α + β + 1 = 2 ⇒ α + β = 1

product = αβ = -6

(α – β)^{2} = (α + β )^{2} – 4αβ = 1 + 24 = 25

α – β = 5

α + β = 1

Adding 2α = 6 ⇒ α = 3

∴ α = 3 and β = -2

Question 10.

If α, β and γ are the roots of x^{3} – 2x^{2} + 3x – 4 = 0, then find

i) Σα^{2}β^{2}

ii) Σαβ (α + β)

Solution:

Since α, β, γ are the roots of

x^{3} – 2x^{2} + 3x – 4 = 0

then α + β + γ = 2

αβ + βγ + γα = 3

αβγ = 4

i) Σα^{2}β^{2} = α^{2}β^{2} + β^{2}γ^{2} + γ^{2}α^{2}

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

= 9 – 2. 2.4 = 9 – 16 = -7

ii) Σαβ (α + β) = α^{2}β + β^{2}γ + γ^{2}α + αβ^{2} + βγ^{2} + γα^{2}

= (αβ + βγ + γα)(α + β + γ) – 3αβγ

= 2.3 – 3.4 = 6 – 12 = -6.

Question 11.

Solve the x^{3} – 3x^{2} – 6x + 8 = 0 equation, given that the roots of each are in A.P. (‘Mar. ’07)

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, -2

∴ The roots are -2, 1, 4.

Question 12.

Solve x^{4} – 4x^{2} + 8x + 35 = 0, given that 2 + i \(\sqrt{3}\) is a root (AP Mar. ’15)

Solution:

Let 2 + i \(\sqrt{3}\) is one root

⇒ 2 – i \(\sqrt{3}\) is another root.

The equation having roots

2 ± i \(\sqrt{3}\) is x^{2} – 4x + 7 = 0

∴ x^{2} – 4x + 7 is a factor of

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

Question 13.

Find the polynomial equation whose roots are the reciprocals of the roots of x^{4} – 3x^{3} + 7x^{2} + 5x – 2 = 0 (TS Mar. ’15, ’11)

Solution:

Given equation is

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

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

i.e. \(\frac{1}{x^{4}}\) – \(\frac{3}{x^{3}}\) + \(\frac{7}{x^{2}}\) + \(\frac{5}{x}\) – 2 = 0

Multiplying with x^{4}

⇒ 1 – 3x + 7x^{2} + 5x^{3} – 2x^{4} = 0

i.e., 2x^{4} – 5x^{3} – 7x^{2} + 3x – 1 = 0

Question 14.

Find the polynomial equation whose roots are the translates of those of x^{5} – 4x^{4} + 3x^{2} – 4x + 6 = 0 by -3. (TS Mar. 16)

Solution:

Given equation is

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

Required equation is f(x + 3) = 0

(x + 3)^{5} – 4(x + 3)^{4} + 3(x + 3)^{2}

Required equation is

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

Question 15.

Show that x^{5} – 5x^{3} + 5x^{2} – 1 = 0 has three equal roots and find this root. (TS Mar. ’17)

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).

⇒ 1 is a root of above equation

(∵ sum of the coefficients is zero)

∴ 1 is the required root.

Question 16.

Solve the 8x^{3} – 36x^{2} – 18x + 81 = 0 equation, given that the roots of each are in AP. (Mar. 04’)

Soluution:

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

Let the roots be a – d, a, a + d

Sum of the roots = a – d + a + a + d

= \(\frac{36}{8}\) = \(\frac{9}{2}\)

i.e., 3a = \(\frac{9}{2}\) ⇒ a = \(\frac{3}{2}\)

∴ (x – \(\frac{3}{2}\)) is a factor of

⇒ 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 –\(\frac{3}{2}\), \(\frac{3}{2}\), \(\frac{9}{2}\)

Question 17.

Solve the 3x^{3} – 26x^{2} + 52x – 24 = 0 equations, given that the roots of each are in GP.

(TS Mar. ’15)

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

⇒ 3x^{3} – 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.

Question 18.

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

remaining roots. (May ’11; Mar. ’05)

Solution:

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

Question 19.

Solve the equation

2x^{5} + x^{4} – 12x^{3} – 12x^{2} + x + 2 = 0 (AP Mar.17, 16; Mar. ‘08, 07)

Solution:

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

This is an odd degree reciprocal equation of first type.

∴ -1 is a root.

Dividing f(x) with x + 1

Dividing f(x) by (x + 1) we get

2x^{4} – x^{3} – 11x^{2} – x + 2 = 0

Dividing by x^{2}

Substituting in (1), required equation is

2(a^{2} – 2) – a – 11 = 0 .

2a^{2} – 4 – a – 11 = 0

2a^{2} – a – 15 = 0

(a – 3)(2a + 5) = 0

a = 3 or \(-\frac{5}{2}\)

Case (i) a = 3

Question 20.

Find the roots of

x^{4} – 16x^{3} + 86x^{2} – 176x + 105 = 0

Solution:

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

Now, f(1) = 1 – 16 + 86 – 176 + 105 = 0

∴ 1 is a root of f(x) = 0

⇒ x – 1 is a factor of f(x)

∴ f(x) = (x – 1) (x^{3} – 15x^{2} + 71x – 105)

= (x – 1) g(x) where

g(x) = x^{3} – 15x^{2} + 71x – 105

g(1) = 1 – 15 + 71 – 105 = -48 ≠ 0

g(2) = -15 ≠ 0

g(3) = 27 – 135 + 213 – 105 = 0

∴ 3 is a root of g(x) =0

⇒ x – 3 is a factor of g(x)

∴ g(x) = (x – 3) (x^{2} – 12x + 35)

= (x – 3) (x – 5) (x – 7)

∴ f(x) = (x – 1) (x – 3) (x – 5) (x – 7)

∴ 1, 3, 5, 7 are the roots of f(x) = 0.

Question 21.

Solve 4x^{3} – 24x^{2} + 23x + 18 = 0, given that the roots of this equation are in arithmetic progression (Mar. ’14; May ’06)

Solution:

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

Now, sum of the roots

a – d + a + a + d = \(\frac{24}{4}\)

3a = 6

a = 2

Product of the roots (a – d) a (a + d) = \(-\frac{18}{4}\)

a(a^{2} – d) = \(-\frac{9}{2}\)

2(4 – d^{2}) = \(-\frac{9}{2}\)

4(4 – d^{2}) = -9

16 – 4d^{2} = -9

4d^{2} = 25

d = ±\(\frac{5}{2}\)

∴ roots are –\(\frac{1}{2}\), 2 and \(\frac{9}{2}\)

Question 22.

Find the polynomial equation whose roots are the squares of the roots of x^{5} + 4x^{3} – x^{2} + 11 = 0 by -3. (Mar. ’06)

Solution:

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

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

The required equation is

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

Question 23.

Solve the equation 6x^{4} – 35x^{3} + 62x^{2} – 35x + 6 = 0. (May. 13)

Solution:

We observe that the given equation is an even degree reciprocal equation of class one. On dividing both sides of the given equation by x^{2}, we get

Then the above equation reduces to

6(y^{2} – 2) – 35y + 62 = 0

i.e., 6y^{2} – 35y + 50 = 0

i.e., (2y – 5)(3y – 10) = 0.

Hence the roots of 6y^{2} – 35y + 50 = 0 are \(\frac{5}{2}\) and \(\frac{10}{3}\).

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

Question 24.

Solve x^{5} – 5x^{4} + 9x^{3} – 9x^{2} + 5x – 1 = 0. (Mar. ’13)

Solution:

Given equation is

x^{5} – 5x^{4} + 9x^{3} – 9x^{2} + 5x – 1 = 0 is a reciprocal equation of odd degree and of class two.

∴ 1 is a root of the given equation.

⇒ (x – 1) is a factor of

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

Question 25.

Form the polynomial equation of degree 3 whose roots are 2, 3 and 6. (Mar. ’02)

Solution:

The required polynomial equation is,

(x – 2) (x – 3)(x – 6) = 0

⇒ x^{3} – 11x^{3} + 36x – 36 = 0

Question 26.

Find the relation between the roots and the coefficients of the cubic equation

3x^{3} – 10x^{2} + 7x + 10 = 0.

Solution:

3x^{3} – 10x^{2} + 7x + 1o = 0 ———- (1)

On.comparing (1) with

ax^{3} + bx^{2} + cx + d = 0,

we have

Question 27.

Write down the relations between the roots and the coefficients of the bi-quadratic equation.

x^{4} – 2x^{3} + 4x^{2} + 6x – 21 = 0

Solution:

Given equation is

x^{4} – 2x^{3} + 4x^{2} + 6x – 21 = 0 —— (1)

On comparing (1) with

ax^{4} + bx^{3} + cx^{2} + dx + c = 0,

we have

Question 28.

If 1, 2, 3 and 4 are the roots of x^{4} + ax^{3} + bx^{2} + cx + d = 0, then find the

values of a, b, c and d.

Solution:

Given that the roots of the given equation are 1, 2, 3 and 4. Then

x^{4} + ax^{3} + bx^{2} + cx + d

≡ (x – 1) (x – 2) (x – 3) (x – 4) = 0

≡ x^{4} – 10x^{3} + 35x^{2} – 50x + 24 = 0

On equating the coefficients of like powers of x, we obtain

a = -10, b = 35, c = -50, d = 24

Question 29.

if a, b, c are the roots of

x^{3} – px^{2} + qx- r = 0 and r ≠ 0, then find \(\frac{1}{\mathbf{a}^{2}}+\frac{1}{\mathbf{b}^{2}}+\frac{1}{\mathrm{c}^{2}}\) interms of p, q, r.

Solution:

Given that a, b, c are the roots of

x^{3} – px^{2} + qx – r = 0, then

a + b + c = p, ab + bc + ca = q, abc = r

Question 30.

Find the sum of the squares and the sum of the cubes of the roots of the equation x^{3} – px^{2} + qx – r = 0 in terms of p, q, r.

Solution:

Let α, β, γ be the roots of the given equation then α + β + γ = p, αβ + βγ + γα = q, αβγ = r

Sum of the squares of the roots is α^{2} + β^{2} + γ^{2}

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

Sum of the cubes of the roots is α^{3} + β^{3} + γ^{3}

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

= p(p^{2} – 2q – q) + 3r

= p(p^{2} – 3q) + 3r

Question 31.

Obtain the cubic equation, whose roots are the sqüares of the roots of the equation, x^{3} + p_{1}x^{2} + P_{2}x + p_{3} = 0

Solution:

The required equation is, f(\(\sqrt{x}\)) =0

Question 32.

Let α, β, γ be the roots of

x^{3} + px^{2} + qx + r = 0. Then find the

i) Σα^{2}

ii) Σ\(\frac{1}{\alpha}\)

iii) Σα^{3}

iv) Σβ^{2}γ^{2}

v) Σ(α + β) (β + γ) (γ + α)

Solution:

since α, β, γ are the roots of the equation, we have α + β + γ = – p,

αβ + βγ + γα = q, αβγ = -r.

i) Σα^{2}

Solution:

Σα^{2} = α^{2} + β^{2} + γ^{2}

= (α + β + γ) – 2(αβ + βγ + γα)

= p^{2} – 2q

ii) Σ\(\frac{1}{\alpha}\)

Solution:

iii) Σα^{3}

Solution:

iv) Σβ^{2}γ^{2}

Solution:

v) (α + β) (β + γ) (γ + α)

Solution:

We know, α + β + γ = -p

⇒ α + β = -p – r and β + γ = -p – α

= γ + α = -p – β

∴ (α + β) (β + γ)(γ + α)

= (-p – γ) (-p – α) (-p – β)

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

= -p^{3} + p^{3} – pq + r = r – pq .

Question 33.

Let α, β, γ be the roots of

x^{3} + ax^{2} + bx + c = 0 then find Σα^{2} + Σβ^{2}

Solution:

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

Question 34.

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

α(β + γ), β(γ + α), γ(α + β)

Solution:

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

we have, α + β + γ = – p, αβ + βγ + γα = q, αβγ = -r

Let y = α(β + γ)

= αβ + αγ + γβ – βγ

Question 35.

Solve x^{3} – 3x^{2} – 16x + 48 = 0

Solution:

Let f(x) = x^{3} – 3x^{2} – 16x + 48

by inspection, f(3) = 0

Hence 3 is a root of 1(x) = 0

Now we divide f(x) by (x – 3)

Question 36.

Find the roots of

x^{4} – 16x^{3} + 86x^{2} – 176x + 105 = 0 (Mar. ‘02)

Solution:

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

Now if (1) = 1 – 16 + 86 – 176 + 105 = 0

∴ 1 is a root of f(x) = 0

⇒ x – 1 is a factor of f(x)

∴ g(x) = (x – 3)(x^{2} – 12x + 35)

= (x – 3) (x – 5) (x – 7)

∴ f(x) = (x – 1) (x – 3) (x – 5) (x – 7)

∴ 1, 3, 5, 7 are the roots of f(x) = 0.

Question 37.

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

Solution:

Let α, β, γ be the root of the equation

x^{3} – 7x^{2} + 36 = 0 and

let β = 2α

Now, we have, α + β + γ = 7

⇒ 3α + γ = 7 —— (1)

αβ + βγ + γα = 0

⇒ 2α^{2} + 3αγ = 0 —— (2)

αβγ = -36 ⇒ 2α^{2}γ = -36 —— (3)

From (1) and (2), we have

2α^{2} + 3α(7 – 3α) = 0

i.e., α^{2} – 3α = 0 (or) α(α – 3) = 0

∴ α = 0 or α = 3

Since α = 0 does not satisfy the given equation.

∴ α = 3, so β = 6 and γ = -2

∴ The roots are 3, 6, -2.

Question 38.

Given that 2 is a root of

x^{3} – 6x^{2} + 3x + 1o = 0, find the other roots.

Solution:

Let f(x) = x^{3} – 6x^{2} + 3x + 10

Since 2 is a root of f(x) = 0, we divide f(x) by (x – 2)

∴ -1, 2, and 5 are the roots of the given equation.

Question 39.

Given that two roots of

4x^{3} + 20x^{2} – 23x + 6 = 0 are equal, find all the roots of the given equation.

Solution:

Let α, β, γ are the roots of

4x^{3} + 20x^{2} – 23x + 6 = 0

Given two roots are equal, let α = β

⇒ 12α^{2} + 4α – 23 = 0

⇒ (2α – 1) (6α + 23) = 0

α = \(\frac{1}{2}\), α = \(\frac{-23}{6}\)

On verfication, we get that is a root of (1)

α = \(\frac{1}{2}\) is roots of (1)

(2) ⇒ roots are \(\frac{1}{2}\), \(\frac{1}{2}\), -6

Question 40.

Given that the sum of two roots of

x^{4} – 2x^{3} + 4x^{2} + 6x – 21 = 0 is zero find the roots of the equation.

Solution:

Let α, β, γ, δ are the roots of given equation, since sum of two is zero.

α + β = 0

Now α + β + γ + δ = 2 ⇒ γ + δ = 2

Let αβ = p, γδ = q

The equation having the roots α, β is

∴ Roots are –\(\sqrt{3}\), \(\sqrt{3}\), 1 – i \(\sqrt{6}\) and 1 + i \(\sqrt{6}\)

Question 41.

Solve 4x^{3} – 24x^{2} + 23x + 18 = 0, given that the roots of this equation are in arithmetic progression. (Mar. 14, May ‘06)

Solution:

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

Now, sum of the roots

a – d + a + a + d = \(\frac{24}{4}\)

3a = 6

a = 2

Product of the roots (a – d) a (a + d) = \(\frac{-18}{4}\)

a(a^{2} – d^{2}) = \(\frac{-9}{2}\)

2(4 – d^{2}) = \(\frac{-9}{2}\)

4(4 – d^{2}) = -9

16 – d^{2} = -9

4d^{2} = 25

d = ± \(\frac{5}{2}\)

∴ roots are –\(\frac{1}{2}\), 2 and \(\frac{9}{2}\)

Question 42.

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

Solution:

Let \(\frac{a}{r}\), a, ar be the roots of the given equation. Then .

Hence a = 2. On substituting a = 2 in (1), we obtain

\(\frac{2}{r}\) + 2 + 2r = 7

i.e., 2r^{2} – 5r + 2 = 0

i.e., (r – 2) (2r – 1) = 0

Therefore r = 2 or r = \(\frac{1}{2}\)

Hence the roots of the given equation are 1, 2 and 4.

Question 43.

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 be the roots of the given equation.

Product of the roots αβγδ = -6

Given αβ = 3 (∵ Product of two roots is 3)

∴ α, β, γ, δ = -6

γδ = -2

Let α + β, γ + δ = q

The equation having the roots α,β is

x^{2} – (α + β) x + αβ = 0

x^{2} + px + 3 = 0

The equation having the roots γ, δ is

x^{2} – (γ + δ)x + γδ = 0

x^{2} – qx – 2 = 0

∴ x^{4} – 5x^{3} + 5x^{2} + 5x – 6

= (x^{2} – px + 3)(x^{2} – qx – 2)

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

Comparing the like terms,

p + q = 5, 2p – 3q = 5

∴ 2p – 3q = 5

3p + 3q = 15

5p = 20 ⇒ p = 4

∴ q = 1

Now x^{2} – 4x + 3 = 0 ⇒ (x – 3)(x – 1) = 0

⇒ x = 1, 3

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

⇒ x = -1, 2

∴ The roots are -1, 2, 1, 3

Question 44.

Solve x^{6} + 4x^{3} – 2x^{2} – 12x + 9 = 0, Given that it has two pairš of equal roots.

Solution:

Given equation is

x^{4} + 4x^{3} – 2x^{2} – 12x + 9 = 0

Let the roots be α, α, β, β

Sum of the roots, 2(α + β) = -4

⇒ α + β = -2

Let αβ = p

The equation having roots α, β is

x^{2} – (α + β)x + αβ = 0

i.e. x^{2} + 2x + p = 0

∴ x^{4} + 4x^{3} – 2x^{2} – 12x + 9

= [x^{2} – (α + β)x + αβ]^{2}

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

= x^{4} + 4x^{3} + (2p + 4)x^{2} + 4px + p^{2}

Comparing coefficients of x on both sides

4p = -12 ⇒ p = – 3

x^{2} + 2x + p = 0 ⇒ x^{2} + 2x – 3 = 0

⇒ (x + 3)(x – 1) = 0

⇒ x = -3, 1

∴ The roots of the given equation are -3, -3, 1, 1

Question 45.

Prove that the sum of any two of the roots of the equation x^{4} + px^{3} + qx^{2} + rx + s = 0 is equal to the sum of the remaining two roots of the equation iff p^{3} – 4pq + 8r = 0.

Solution:

Suppose that the sum of two of the roots of the given equation is equal to the sum of the remaining two roots.

Let α, β, γ, δ be roots of the given equation such that α + β = γ + δ

From these equations, we have

Then equations (1), (2) and (4) are satisfied. In view of (5), equation (3) is also satisfied. Hence (x^{2} + bx + c)(x^{2} + bx + d) = x^{4} + 2bx^{3} + (b^{2} + c + d)x^{2} + b(c + d)x + cd = x^{4} + px^{3} + qx^{2} + rx + s

Hence the roots of the given equation are α_{1}, β_{1}, γ_{1} and δ_{1}. where α_{1} and β_{1} are the roots of the equations x^{2} + bx + c = 0 and γ_{1} and δ_{1} are those of the equation x^{2} + bx + d = 0.

We have α_{1} + β_{1} = -b = γ_{1} + δ_{1}.

Question 46.

Form the polynomial equation of degree 4 whose roots are

4 + \(\sqrt{3}\), 4 – \(\sqrt{3}\), 2 + i and 2 – i

Solution:

The equation having roots 4 + \(\sqrt{3}\), 4 – \(\sqrt{3}\) is

x^{2} – 8x + 13 = 0

The equation having roots 2 + i, 2 – i is

x^{2} – 4x + 5 = 0.

The required equation is

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

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

Question 47.

Solve 6x^{4} – 13x^{3} – 35x^{2} – x + 3 = 0 given that one of its root is 2 + \(\sqrt{3}\).

Solution:

2 + \(\sqrt{3}\) is a root 2 – \(\sqrt{3}\) is also a root.

The equation having roots

Question 48.

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

The required equation is f(-x) = 0

i.e., (-x)^{4} – 6(-x)^{3} + 7(-x)^{2} + 2(-x) + 1 = 0

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

Question 49.

Find the algebraic equation of the 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

The required equation is f\(\left(\frac{x}{3}\right)\) = o

Question 50.

Form the equation whose roots are m times the roots of the equation x^{3} + \(\frac{x^{2}}{4}\) – \(\frac{x}{16}\) + \(\frac{1}{72}\) = 0 and deduce the case when m = 12.

Solution:

Question 51.

Find the algebraic equation of degree 5 whose roots are the translates of the roots of x^{5} + 4x^{3} – x^{2} + 11 = 0 by -3. (Mar. ’06)

Solution:

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

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

The required equation is

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

Question 52.

Find the algebraic equation of degree 4 whose roots are the translates of the roots

4x^{4} + 32x^{3} + 83x^{2} + 76x + 21 = 0 by 2.

Solution:

Let f(x) ≡ 4x^{4} + 32x^{3} + 83x^{2} + 76x + 21

The required equation is f(x – 2) = 0

The required equation is

4x^{4} – 13x^{2} + 9 = 0

Question 53.

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

Question 54.

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 .

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

Squaring on both sides

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

= x^{3} + 6x^{2} + 64x – x^{2} – 12x – 36 = 0

∴ x^{3} + 15x^{2} + 52x – 36 = 0

Question 55.

Show that 2x^{3} + 5x^{2} + 5x + 2 = 0 is a reciprocal equation of class one.

Solution:

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

P_{0} = 2, p_{1} = 5, P_{2} = 5, p_{3} = 2

Here P_{0} = p_{3}, p_{1} = p_{2}

∴ The equation 2x^{3} + 5x^{2} + 5x + 2 = 0 is a reciprocal equation of class one.

Question 56.

Solve the equation

4x^{3} – 13x^{2} – 13x + 4 = 0

Solution:

4x^{3} – 13x^{2} – 13x + 4 = 0 is a reciprocal equation of first class and of odd degree.

Thus -1 is a root of the 9iven equation.

4x^{2} – 17x + 4 = 0 ⇒ 4x^{2} – 16x – x + 4 = 0

⇒ 4x(x – 4) – 1 (x – 4) = 0

⇒ (x – 4)(4x – 1) = 0

⇒ x = 4 or \(\frac{1}{4}\)

The roots are -1, 4, \(\frac{1}{4}\)

Question 57.

Solve the equation 6x^{4} – 35x^{3} + 62x^{2} – 35x + 6 = 0. (May ‘13)

Solution:

We observe that the given equation is an even degree reciprocal equation of class one.

On dividing both sides 6f the given equation by x^{2}, we get

Then the above equation reduces to

6(y^{2} – 2) – 35y + 62 = 0

i.e., 6y^{2} – 35y + 50 = 0

i.e., (2y – 5) (3y – 10) = 0.

Hence the roots of 6y^{2} – 35y + 50 = 0 are \(\frac{5}{2}\) and \(\frac{10}{3}\).

Therefore x + \(\frac{1}{x}\) = \(\frac{5}{2}\) and x + \(\frac{1}{x}\) = \(\frac{10}{3}\)

i.e., 2x^{2} – 5x + 2 = 0 and 3x^{2} – 10x + 3 = 0.

The roots of these equations are respectively

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

Question 58.

Solve x^{5} – 5x^{4} – 9x^{3} – 9x^{2} + 5x – 1 = 0. (Mar ’13)

Solution:

Given equation is

x^{5} – 5x^{4} + 9x^{3} – 9x^{2} + 5x – 1 = 0 is a reciprocal equation’of odd degree and of class two.

∴ 1 is a root of the given equation.

⇒ (x – 1) is a factor of

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

Question 59.

Solve the equation

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

Solution:

Given equation is

6x^{6} – 25x^{5} + 31x^{4} – 31x^{2} + 25x – 6 = 0 is a reciprocal equation of second class and of even degree.

∴ x^{2} – 1 is a factor of

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

∴ (1) becomes 6(y^{2} – 2) – 25(y) + 37 = 0

⇒ 6y^{2} – 12 – 25y + 37 = 0

⇒ 6y^{2} – 25y + 25 = 0

⇒ 6y^{2} – 15y – 10y + 25 = 0

⇒ 3y(2y – 5) – 5(2y – 5) = 0

⇒ (2y – 5)(3y – 5) = 0