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

## TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d)

I.

Question 1.

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

Solution:

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

Let f(x) = x^{3} + 2x^{2} – 4x + 1

The equation whose roots are 3 times the roots of f(x) = 0 is given by f(\(\frac{x}{3}\)) = 0.

i.e., \(\left(\frac{x}{3}\right)^3+2\left(\frac{x}{3}\right)^2-4\left(\frac{x}{3}\right)\) + 1 = 0

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

⇒ x^{3} + 6x^{2} – 36x + 27 = 0.

Question 2.

Find the algebraic equation whose roots are 2 times the roots of x5 – 2×4 + 3×3 – 2×2 + 4x + 3 = 0.

Solution:

Given equation is x^{5} – 2x^{4} + 3x^{3} – 2x^{2} + 4x + 3 = 0 ……..(1)

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

The equation whose roots are 2 times the roots of f(x) = 0 is given by f(\(\frac{x}{2}\)) = 0

i.e., \(\left(\frac{x}{2}\right)^5-2\left(\frac{x}{2}\right)^4+3\left(\frac{x}{2}\right)^3-2\left(\frac{x}{2}\right)^2+4\left(\frac{x}{2}\right)+3\) = 0

⇒ x^{5} – 4x^{4} + 12x^{3} – 16x^{2} + 64x + 96 = 0.

Question 3.

Find the transformed equation whose roots are the negatives of the roots of x4 + 5×3 + lix + = 0.

Solution:

Given equation is x^{4} + 5x^{3} + 11x + 3 = 0

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

The transformed equation whose roots are the negatives of the roots of f(x) = 0 is

f(- x) = 0.

i.e., (- x)^{4} + 5(- x)^{3} + 11 (- x) + 3 = 0

⇒ x^{4} – 5x^{3} – 11x + 3 = 0.

Question 4.

Find the transformed equation whose roots are the negatives of the root of x^{7} + 3x^{5} + x^{3} – x^{2} + 7x + 2 = 0.

Solution:

Given equation is

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

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

The transformed equation whose roots are the negatives of the roots of f(x) = 0 is

f(- x) = 0.

i.e., (- x)^{2} + 3(- x)^{5} + (- x)^{3} – (- x)^{2} + 7(- x) + 2 = 0

x^{7} + 3x^{5} + x^{3} + x^{2} + 7x – 2 = 0.

Question 5.

Find the polynomial equation whose roots are the reciprocals of the roots of x^{4} – 3x^{3} + 7x^{2} + 5x – 2 = 0.

Solution:

Given equation is x^{4} – 3x^{3} + 7x^{2} + 5x – 2 = 0 ………….(1)

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

The polynomial equation whose roots are the reciprocals of the roots of (1) is given by

f(\(\frac{1}{x}\)) = 0

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

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

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

Question 6.

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:

Given equation is x^{5} + 11x^{4} + x^{3} + 4x^{2} – 13x + 6 = 0 …………(1)

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

The polynomial equation whose roots are the reciprocals of the roots of f(x) = 0 is

f(\(\frac{1}{x}\)) = 0

i.e., \(\left(\frac{1}{x}\right)^5+11\left(\frac{1}{x}\right)^4+\left(\frac{1}{x}\right)^3\) + 6 = 0

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

II.

Question 1.

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

Solution:

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

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

The polynomial equation whose roots are squares of the roots of f(x) = 0 is f (√x) = 0.

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

⇒ x^{2} + 2x + 1 = √x (x + 1)

⇒ (x + 1)^{2} = √x (x + 1)

⇒ (x + 1)^{4} = x (x + 1)^{2}

⇒ x^{4} + 4x^{3} + 6x^{2} + 4x + 1 = x(x^{2} +2x + 1)

⇒ x^{4} + 3x^{3} + 4x^{2} + 3x + 1 = 0 is the required equation.

Question 2.

Form the polynomial equation whose roots are the squares of the roots of x^{3} + 3x^{2} – 7x + 6 = 0.

Solution:

Given equation is x^{3} + 3x^{2} – 7x + 6 = 0 …… (1)

Let f(x) = x^{3} + 3x^{2} – 7x + 6

The polynomial equation whose roots are the squares of the roots of f(x) = 0 is f(√x) = 0.

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

⇒ x√x – 7√x = – 3x – 6

⇒ √x (x – 7) = – (3x + 6)

⇒ x (x^{2} + 49 – 14x) = 9x^{2} + 36 + 36x

⇒ x^{3} – 23x^{2} + 13x – 36 = 0 is the required equation.

Question 3.

Form the polynomial equation whose roots are the cubes of the roots of x^{3} + 3x^{2} + 2 = 0.

Solution:

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

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

The polynomial equation whose roots are the cubes of the roots of f(x) = 0 is f(\(\sqrt[3]{x}\)) = 0.

i.e., \((\sqrt[3]{x})^3\) + 3 (\((\sqrt[3]{x})^2\)) + 2 = 0

⇒ x + 2 = – 3x^{2/3}

⇒ (x + 2)^{3} = – 27 (x^{2})

⇒ x^{3} + 6x^{2} + 12x + 8 = – 27x^{2}

⇒ x^{3} + 33x^{2} + 12x + 8 = 0 is the required equation.

III.

Question 1.

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

Solution:

Given equation is

x^{4} – 5x^{3} + 7x^{2} – 17x + 11 = 0 ………(1)

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

The polynomial equation, whose roots are

the translates of those of the f(x) = 0 by – 2 is f(x + 2) = 0.

Suppose that

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

By synthetic division, the coefficients A_{0}, A_{1}, A_{2}, A_{3} and A_{4} are obtained as follows.

∴ The roots ol the equation x^{4} – 3x^{3} + x^{2} – 17x + 19 = 0 are the translates of the roots of the given equation by – 2.

Question 2.

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

Solution:

Given equation is x_{5} – 4x_{4} + 3x_{2} – 4x 4 6 = 0 ………….(1)

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

The polynomial equation, whose roots are the translates of those of the f(x) = 0 by – 3 is f(x + 3) = 0.

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

By synthetic division, the coefficients A_{0}, A_{1}, A_{2}, A_{3}, A_{4} and A_{5} are obtained as follows.

∴ The roots of the equation x^{5} + 11x^{4} + 42x^{3} + 57x^{2} – 13x – 60 = 0 are the translates of the roots of the given equation by – 3.

Question 3.

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:

Given equation is x^{4} – x^{3} – 10x^{2} + 4x + 24 = 0

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

The polynomial equation whose roots are trans-lates of those of the f(x) = 0 by 2 is f(x – 2) = 0.

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

By synthetic division, the coefficients A_{0}, A_{1}, A_{2}, A_{3}, A_{4} are obtained as follows – 2.

∴ The roots of the equation x^{4} – 9x^{3} + 20x^{2} = 0 are the translates of the roots of the given equati on by 2.

Question 4.

Find the polynomial equation whose roots are the translates of the roots of the equation 3x^{5} – 5x^{3} + 7 = 0 by 4.

Solution:

Given equation is 3x^{5} – 5x^{3} + 7 = 0

Let f(x) = 3x^{5} – 5x^{3} + 7

The polynomial equation whose roots are the translates of those of the f(x) = 0 by 4 is f(x – 4) = 0.

Suppose that

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

By synthetic division, the coefficients A_{0}, A_{1}, A_{2}, A_{3}, A_{4}, A_{5} are obtained as follows.

∴ The roots of the equation 3x^{5} – 60x^{4} + 475x^{2} – 1860x^{3} + 3600x – 2745 = 0 are the translates of the roots of the given equation by 4.

Question 5.

Transform e jach of the following equations into ones i n which the coefficients of the second hig >hest power of x is zero and also find their transformed equations.

i) x^{3} – 6x^{2} + 10x – 3 = 0

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

iii) x^{3} – 6x^{2} + 4x – 7 = 0

iv) x^{3} + 6x^{2}+ 4x + 4 = 0

Solution:

i) Given equation is x^{3} – 6x + 10x – 3 = 0

Let f(x) = x^{3} – 6x^{2} + 10x – 3

We have to find ’h’ so that the coefficient of the Second highest power of x in f(x + h) is zero.

i.e., Coefficient of x^{2} in f(x + h) is zero,

f (x + h) = (x + h)^{3} – 6 (x + h)^{2} + 10 (x + h) – 3 Coefficients of x^{2} in f(x + h) is 3h – 6.

We choose ‘h‘ such that 3h – 6 = 0 i.e., h = 2

∴ f (x + 2) = (x + 2)^{3} – 6 (x + 2)^{2} + 10 (x + 2) – 3

= x^{3} + 6x^{2} + 12x + 8 – 6 (x^{2} + x + 4) + 10x + 20 – 3

= x^{3} – 2x + 1

∴ x^{3} – 2x + 1 = 0 is the required equation.

ii) Given equation is

x^{4} + 4x^{3} + 2x^{2} – 4x – 2 = 0

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

We have to find ‘h’ so that the coefficient of the second highest power of x in f(x + h) is zero.

i.e., coefficient of x^{2} in f(x + h) is zero,

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

Coefficient of x^{2} in f(x + h) is 4h + 4.

We choose ‘h‘ such that 4h + 4 = 0 i.e., h = – 1

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

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

= x^{4} – 4x^{2} + 1

∴ x^{4} – 4x^{2} + 1 = 0 is the required equation.

iii) Given equation is x^{3} – 6x^{2} + 4x – 7 = 0

Let f(x) = x^{3} – 6x^{2} + 4x – 7

We have to find ‘h’ so that the coefficient of the second highest power of x in f(x + h) is zero.

i.e., coefficient of x^{2} in f(x + h) is zero.

f(x + h) = (x + h)^{3} – 6 (x + h)^{2} + 4 (x + h) – 7

Coefficient of x^{2} in f(x + h) is 3h – 6

We choose ‘h’ such that 3h – 6 = 0 i.e., h = 2

∴ f(x + 2) = (x + 2)^{3} – 6 (x + 2)^{2} + 4 (x + 2) – 7

= (x^{3} + 6x^{2} + 12x + 8) – 6 (x^{2} + 4x + 4) + 4 (x + 2) – 7

= x^{3} – 8x – 15

∴ x3 – 8x – 15 = 0 is the required equation.

iv) Given equation is x^{3} + 6x^{2} + 4x + 4 = 0

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

We have to find ’h’ so that the coefficient of the second highest power of x in f(x + h) is zero, i.e., coefficient of x^{2} in f(x + h) is zero.

f(x + h) = (x + h)^{3} + 6(x + h)^{2} + 4(x + h) + 4

Coefficient of x^{2} in 1(x + h) is 3h + 6.

We have to choose ‘h’ such that 3h + 6 = 0 i.e., h = – 2

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

= (x^{3} – 6x^{2} + 12x – 8) + 6 (x^{2} – 4x + 4) + 4 (x – 2) + 4

= x^{3} – 8x + 12

∴ x^{3} – 8x + 12 = is the required equation.

Question 6.

Transform each of the following equations into ones in which the coefficients of the third highest power of x is zero.

i) x^{4} + 2x<sup3 – 12x^{2} + 2x – 1 = 0

ii) x^{3} + 2x^{2} + x + 1 = 0

Solution:

i) Given equation is

x^{4} + 2x^{3} – 12x^{2} + 2x – 1 = 0

Let f(x) = x^{4} + 2x^{3} – 12x^{2} + 2x – 1

We have to find h’ so that the coefficient of the third highest power of ‘x’ in f(x + h) is zero.

i.e., Coefficient of x^{2} in [(x + h) is zero.

f(x + h) = (x + h)^{4} + 2 (x + h)^{3} – 12 (x + h)^{2} + 2 (x + h) – 1

Coefficient of x^{3} in 1(x + h) is 6h^{2} + 6h – 12

We have to choose ‘h’ such that

6h^{2} + 6h – 12 = 0

⇒ (h + 2) (h – 1) = 0

⇒ h = – 2 or 1.

Case – (I):

When h = – 2

f(x – 2) = (x – 2)^{4} + 2 (x – 2)^{3} – 12(x – 2)^{2} + 2 (x – 2) – 1

= x^{4} – 8x^{3} + 24x^{2} – 32x + 16 + 2(x^{3} – 6x^{2} + 12x – 8) – 12(x^{2} – 4x + 4) + 2(x – 2) – 1

= x^{4} – 6x^{3} + 42x – 53

∴ Tranformed equation is x^{4} – 6x^{3} + 42x – 53 = 0.

Case-(ii):

When h = 1

f(x + 1) = (x + 1)^{4} + 2(x + 1)^{3} – 12(x + 1)^{2} + 2 (x + 1) – 1

= (x^{4} + 4x^{3} + 6x^{2} + 4x + 1) + 2(x^{3} + 3x^{2} + 3x + 1) – 12 (x^{2} + 2x + 1) + 2(x + 1) – 1

= x^{4} + 6x^{3} – 12x – 8

∴ Tranformed equation is x^{4} + 6x^{3} – 12x – 8 = 0

∴ Required equation is x^{4} – 6x^{3} + 4x – 53 = 0

or x^{4} + 6x^{3} – 12x – 8 = 0.

ii) Given equation is x^{3} + 2x^{2} + x + 1 = 0

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

We have o find ‘h’ so that the coefficient of the third highest power of ‘x’ in f(x + h) is zero.

i.e., Coefficient of x in f(x + h) is zero.

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

Coefficient of ‘x^{3}’ in f(x + h) is 3h^{2} + 4h + 1

We have to Choose ‘h’ such that 3h^{2} + 4h + 1 = 0

i.e., h = – 1 or h = – \(\frac{1}{3}\).

Case – (I):

When h = – 1

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

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

= x^{3} – x^{2} + 1

∴ Transformed euluation is x^{3} – x^{2} + 1 = 0.

Case – (ii):

When h = – \(\frac{1}{3}\)

\(f\left(x-\frac{1}{3}\right)=\left(x-\frac{1}{3}\right)^3+2\left(x-\frac{1}{3}\right)^2+\left(x-\frac{1}{3}\right)\)

= \(\left(x^3-x^2+\frac{x}{3}-\frac{1}{27}\right)+2\left(x^2-\frac{2}{3} x+\frac{1}{9}\right)\) + x – \(\frac{1}{3}\) + 1

= x^{3} + x^{2} + \(\frac{23}{27}\)

∴ Transformed equation is x^{3} + x^{2} + \(\frac{23}{27}\)

⇒ 27x^{3} + 27x^{2} + 23 = 0

∴ Required equation is x^{3} – x^{2} + 1 = 0 or 27x^{3} + 27x^{2} + 23 = 0.

Question 7.

Solve the following equations.

i) x^{4} – 10x^{3} + 26x^{2} – 10x + 1 = 0.

ii) 2x^{5} + x^{4} – 12x^{3} – 12x^{2} + x + 2 = 0

Solution:

i) Given equatIon is

x^{4} – 10x^{3} + 26x^{2} – 10x + 1 = 0 (1)

This is even degree reciprocal equation of class one.

On dividing (1) by 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

⇒ \(\left(x^2+\frac{1}{x^2}\right)-10\left(x+\frac{1}{x}\right)\) + 24 = 0

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

⇒ y^{2} – 10y + 24 = 0

⇒ (y – 4) (y – 6) = 0

⇒ y = 4 or y = 6.

Case – (i):

When y = 4

⇒ x + \(\frac{1}{x}\) = 4

⇒ x^{2} – 4x + 1 = 0

⇒ x = 2 ± √3.

Case – (ii):

When y = 6

⇒ x + \(\frac{1}{x}\) = 4

⇒ x^{2} – 6x + 1 = 0

⇒ x = 3 ± 2√2

∴ The roots are 2 ± √3, 3 ± 2√2.

ii) Given equation is

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

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

(1) is an odd degree reciprocal equation of class one.

∴ x = – 1 is a root of (1)

x + 1 is a factor of f(x).

We divide f(x) with x + 1.

By synthetic division,

∴ f(x) = (x + 1) (2x^{4} – x^{3} – 11x^{2} – x + 2)

Let g(x) = 2x^{4} – x^{3} – 11x^{2} – x + 2

g(x) = 0 is an even degree reciprocal equation 0f class one.

Dividing g(x) = 0 by x^{2}

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

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

⇒ \(2\left(x+\frac{1}{x}\right)^2-\left(x+\frac{1}{x}\right)\) – 15 = 0

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

⇒ 2y^{2} – y – 15 = 0

⇒ (2y + 5) (y – 3) = 0

∴ y = 3 or y = – 1.

Case – (i) :

When y = 3

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

⇒ x^{2} – 3x + 1 = 0

⇒ x = \(\frac{3 \pm \sqrt{5}}{2}\)

Case – (ii):

When y = \(\frac{-5}{2}\)

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

⇒ 2x^{2} + 2 = – 5x

⇒ 2x^{2} + 5x + 2 = 0

⇒ (2x + 1) (x + 2) = 0

∴ x = – \(\frac{-1}{2}\) or x = – 2

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