Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Theory of Equations Solutions Exercise 4(d) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2A Theory of Equations Solutions Exercise 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 f(x) = x^{3} + 2x^{2} – 4x + 1 = 0

We require an equation whose roots are 3 times the roots of f(x) = 0

i,e., Required equation is f(\(\frac{x}{3}\)) = 0

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

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

Multiplying with 27, required equation is x^{3} + 6x^{2} – 36x + 27 = 0

Question 2.

Find the algebraic equation whose roots are 2 times the roots of x^{5} – 2x^{4} + 3x^{3} – 2x^{2} + 4x + 3 = 0

Solution:

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

We require an equation whose roots are 2 times the roots of f(x) = 0

Required equation is f(\(\frac{x}{2}\)) = 0

⇒ \(\left(\frac{x}{2}\right)^{5}-2\left(\frac{x}{2}\right)^{4}+3\left(\frac{x}{2}\right)^{3}-2\left(\frac{x}{2}\right)^{2}+4\left(\frac{x}{2}\right)\) + 3 = 0

⇒ \(\frac{x^{5}}{32}-2 \cdot \frac{x^{4}}{16}+3 \cdot \frac{x^{3}}{8}-2 \cdot \frac{x^{2}}{4}+4 \cdot \frac{x}{2}+3=0\)

Multiplying with 32, the required equation is

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

Question 3.

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 4.

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

Solution:

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

We want an equation whose roots are -α_{1}, -α_{2}, ………., -α_{n}

Required equation is f(-x) = 0

⇒ (-x)^{7} + 3(-x)^{5} + (-x)^{3} – (-x)^{2} + 7(-x) + 2 = 0

⇒ -x^{7} – 3x^{5} – x^{3} – x^{2} – 7x + 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 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

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

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

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

Multiplying by x^{5}

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

Required equation f(√x) = 0

⇒ x^{2} + x√x + 2x + √x + 1 = 0

⇒ √x(x + 1) = -(x^{2} + 2x + 1)

Squaring both sides,

⇒ x(x + 1)^{2} = (x^{2} + 2x + 1)^{2}

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

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

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

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 f(x) = x^{3} + 3x^{2} – 7x + 6 = 0

Required equation is f(√x) = 0

⇒ x√x + 3x – 7√x + 6 = 0

⇒ √x(x – 7) = -(3x + 6)

Squaring on both sides,

⇒ x(x – 7)^{2} = (3x + 6)^{2}

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

⇒ x^{3} – 14x^{2} + 49x – 9x^{2} – 36x – 36 = 0

⇒ x^{3} – 23x^{2} + 13x – 36 = 0

Question 3.

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

Solution:

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

Put y = x^{3} so that x = y^{1/3}

∴ y + 3y^{2/3} + 2 = 0

∴ 3y^{2/3} = -(y + 2)

Cubing on both sides,

27y^{2} = -(y + 2)^{3} = -(y^{3} + 6y^{2} + 12y + 8)

∴ y^{3} + 6y^{2} + 12y + 8 = 0

⇒ y^{3} + 33y^{2} + 12y + 8 = 0

Required equation is x^{3} + 33x^{2} + 12x + 8 = 0

III.

Question 1.

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.

Solution:

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

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

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

Required equation is x^{4} + 3x^{3} + x^{2} – 17x – 19 = 0

Question 2.

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

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

Required equation is f(x – 2) = 0

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

Required equation is x^{4} – 9x^{3} + 20x^{2} = 0

Question 4.

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

Solution:

Given f(x) = 3x^{5} – 5x^{3} + 7 = 0

Required equation is f(x – 4) = 0

3(x – 4)^{5} – 5(x – 4)^{3} + 7 = 0

Required equation is x^{5} – 60x^{4} + 475x^{3} – 1860x^{2} + 3600x – 2745 = 0

Question 5.

Transform each of the following equations into ones in which of the coefficients of the second highest power of x is zero and also find their transformed equations.

(i) x^{3} – 6x^{2} + 10x – 3 = 0

Solution:

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

To remove the second term diminish the roots by \(-\frac{a_1}{n a_0}=\frac{6}{3}=2\)

Required equation is x^{3} – 2x + 1 = 0

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

Solution:

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

Diminishing the roots by \(-\frac{a_1}{n a_0}=\frac{-4}{4}=-1\)

Required equation is x^{4} – 4x^{2} + 1 = 0

(iii) x^{3} – 6x^{2} + 4x – 7 = 0

Solution:

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

Diminishing the roots by \(-\frac{a_1}{n a_0}=\frac{6}{3}=2\)

Required equation is x^{3} – 8x – 15 = 0

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

Solution:

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

To remove the second term diminish the roots by \(\frac{-a_1}{n a_0}=-\frac{6}{3}=-2\)

Required equation is x^{3} – 8x + 12 = 0

Question 6.

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

Hint: To remove the r^{th} term in an equation f(x) = 0 of degree n diminish the roots by ‘h’ such that \(f^{(n-r+1)}(h)=0\)

(i) x^{4} + 2x^{3} – 12x^{2} + 2x – 1 = 0

Solution:

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

To remove the 3^{rd} term, diminish the roots by h such that f”(h) = 0

f'(x) = 4x^{3} + 6x^{2} – 24x + 2

f”(x) = 12x^{2} + 12x – 24

f”(h) = 0

⇒ 12h^{2} + 12h – 24 = 0

⇒ h^{2} + h – 2 = 0

⇒ (h + 2) (h – 1) = 0

⇒ h = -2 or 1

Case (i):

Transformed equation is x^{4} – 6x^{3} + 42x – 53 = 0

Case (ii):

Transformed equation is x^{4} – 6x^{3} – 12x – 8 = 0

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

or x^{4} + 6x^{3} – 12x – 8 = 0

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

Solution:

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

To remove the 3^{rd} term, diminish the roots by h such that f'(h) = 0, f'(x) = 3x^{2} + 4x + 1

f'(h) = 0

⇒ 3h^{2} + 4h + 1 = 0

⇒ (3h + 1) (h + 1)

⇒ h = -1, \(-\frac{1}{3}\)

Case (i):

Transformed equation is x^{3} – x^{2} + 1 = 0

Case (ii):

Transformed equation is x^{3} + x^{2} + \(\frac{23}{27}\) = 0

⇒ 27x^{3} + 27x^{2} + 23 = 0

∴ The 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

Solution:

This is a standard reciprocal equation.

Dividing with x^{2}

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

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

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

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

Substituting in (1)

a^{2} – 2 – 10a + 26 = 0

⇒ a^{2} – 10a + 24 = 0

⇒ (a – 4)(a – 6) = 0

⇒ a = 4 or 6

Case (i): a = 4

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

⇒ x^{2} + 1 = 4x

⇒ x^{2} – 4x + 1 = 0

⇒ x = \(\frac{4 \pm \sqrt{16-4}}{2}=\frac{4 \pm 2 \sqrt{3}}{2}\)

⇒ x = 2 ± √3

Case (ii): a = 6

x + \(\frac{1}{x}\) = 6

⇒ x^{2} + 1 = 6x

⇒ x^{2} – 6x + 1 = 0

⇒ x = \(\frac{6 \pm \sqrt{36-4}}{2}\)

⇒ x = \(\frac{2(3 \pm 2 \sqrt{2)}}{2}\)

⇒ x = 3 ± 2√2

∴ The roots are 3 ± 2√2, 2 ± √3

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

Solution:

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

This is an odd-degree reciprocal equation of the 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}

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

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

Put a = x + \(\frac{1}{x}\) so that

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

Substituting in (1), the 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

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

⇒ x^{2} + 1 = 3x

⇒ x^{2} – 3x + 1 = 0

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

Case (ii): a = \(-\frac{5}{2}\)

⇒ \(x+\frac{1}{x}=-\frac{5}{2}\)

⇒ \(\frac{x^2+1}{x}=-\frac{5}{2}\)

⇒ 2x^{2} + 2 = -5x

⇒ 2x^{2} + 5x + 2 = 0

⇒ (2x + 1) (x + 2) = 0

⇒ x = \(-\frac{1}{2}\), -2

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