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

## Intermediate 2nd Year Maths 2A Theory of Equations Solutions Exercise 4(c)

I.

Question 1.

Form the polynomial equation whose roots are

(i) 2 + 3i, 2 – 3i, 1 + i, 1 – i

Solution:

The required equation is [x – (2 + 3i)] [x – (2 – 3i)] [x – (1 + i)][x – (1 – i)] = 0

⇒ [(x – 2) – 3i)] [(x – 2) + 3i] [(x – 1) – i] [(x – 1) + i] = 0

⇒ [(x – 2)^{2} – 9i^{2}] [(x – 1)^{2} – i^{2}] = 0

⇒ (x^{2} – 4x + 4 + 9) (x^{2} – 2x + 1 + 1) = 0

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

⇒ x^{4} – 4x^{3} + 13x^{2} – 2x^{3} + 8x^{2} – 26x + 2x^{2} – 8x + 26 = 0

⇒ x^{4} – 6x^{3} + 23x^{2} – 34x + 26 = 0

(ii) 3, 2, 1 + i, 1 – i

Solution:

Required equation is (x – 3) (x – 2) [x – (1 + i)] [ x – (1 – i)] = 0

⇒ (x^{2} – 5x + 6) [(x – 1) – i] [(x – 1) + i] = 0

⇒ (x^{2} – 5x + 6) [(x – 1)^{2} – i^{2}] = 0

⇒ (x^{2} – 5x + 6) (x^{2} – 2x + 1 + 1) = 0

⇒ (x^{2} – 5x + 6) (x^{2} – 2x + 2) = 0

⇒ x^{4} – 5x^{3} + 6x^{2} – 2x^{3} + 10x^{2} – 12x + 2x^{2} – 10x + 12 = 0

⇒ x^{4} – 7x^{3} + 18x^{2} – 22x + 12 = 0

(iii) 1 + i, 1 – i, -1 + i, -1 – i

Solution:

Required equation is [x – (1 + i)] [x – (1 – i)] [x – (-1 + i)] [x – (-1 – i)] = 0

⇒ [(x – 1) – i][(x – i) + i] [(x + 1) – i] [(x + 1) + i] = 0

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

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

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

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

⇒ x^{4} + 4 = 0

(iv) 1 + i, 1 – i, 1 + i, 1 – i

Solution:

Required equation is [x – (1 + i)] [x – (1 – i)] [x – (1 + i)] [x – (1 – i)] = 0

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

⇒ [(x – 1)^{2} – i^{2}] = 0

⇒ (x^{2} – 2x + 1 + 1)^{2} = 0

⇒ x^{4} + 4x^{2} + 4 – 4x^{3} + 4x^{2} – 8x = 0

⇒ x^{4} – 4x^{3} + 8x^{2} – 8x + 4 = 0

Question 2.

Form the polynomial equation with rational coefficients whose roots are

(i) 4√3, 5 + 2i

Solution:

For the polynomial equation with rational coeffs. the roots are conjugate surds and conjugate complex numbers.

4√3, 5 + 2i

Let α = 4√3 then β = -4√3, and γ = 5 + 2i then δ = 5 – 2i

α, β, γ, δ are the roots

α + β = 0, αβ = -48

γ + δ = 10, γδ = 25 + 4 = 29

The required equation is [x^{2} – (α + β)x + αβ] [x^{2} – (γ + δ)x + γδ] = 0

⇒ (x^{2} – 48) (x^{2} – 10x + 29) = 0

⇒ x^{4} – 10x^{3} + 29x^{2} – 48x^{2} + 480x – 1932 = 0

⇒ x^{4} – 10x^{3} – 19x^{2} + 480x – 1932 = 0

(ii) 1 + 5i, 5 – i

Solution:

For the polynomial equation with rational coeffs. the roots are conjugate surds and conjugate complex numbers.

Let α = 1 + 5i then β = 1 – 5i,

and γ = 5 + i then δ = 5 – i

α + β = 2, αβ = 26

γ + δ = 10, γδ = 26

The required equation is [x^{2} – (α + β)x + αβ] [x^{2} – (γ + δ)x + γδ] = 0

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

⇒ x^{4} – 12x^{3} + 72x^{2} – 312x + 676 = 0

(iii) i – √5

Solution:

For the polynomial equation with rational coeffs. the roots are conjugate surds and conjugate complex numbers.

Let α = i – √5, β = i + √5, γ = -i – √5, δ = -i + √5 are the roots

α + β = 2i, αβ = -6

γ + δ = -2i, γδ = -6

The required equation is [x^{2} – (α + β)x + αβ][x^{2} – (γ + δ)x + γδ] = 0

⇒ (x^{2} – 2ix – 6) (x^{2} + 2ix – 6) = 0

⇒ [(x^{2} – 6) – 2ix] [(x^{2} – 6) + 2ix] = 0

⇒ (x^{2} – 6)^{2} + 4x^{2} = 0

⇒ x^{4} + 36 – 12x^{2} + 4x^{2} = 0

⇒ x^{4} – 8x^{2} + 36 = 0

(iv) -√3 + i√2

Solution:

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

α + β = -2√3

αβ = (-√3)^{2} – (i√2)^{2}

= 3 – i^{2} (2)

= 5

γ + δ = 2√3

γδ = 5

The required equation is [(x^{2} – (α + β)x + αβ)] [(x^{2} – (γ + δ)x + γδ)] = 0

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

⇒ (x^{2} + 5)^{2} – (2√3x)^{2} = 0

⇒ x^{4} + 25 + 10x^{2} – 12x^{2} = 0

⇒ x^{4} – 2x^{2} + 25 = 0

II.

Question 1.

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

Solution:

Let 1 + i be one root ⇒ 1 – i be another root

The equation having roots 1 ± i is x^{2} – 2x + 2 = 0

∴ x^{2} – 2x + 2 is a factor of x^{4} + 2x^{3} – 5x^{2} + 6x + 2 = 0

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

Question 2.

Solve the equation 3x^{3} – 4x^{2} + x + 88 = 0 which has 2 – √-7 as a root.

Solution:

Let 2 – √-7 (i.e) 2 – √7i is one root

⇒ 2 + √7i is another root.

The equation having roots 2 ± √7i is x^{2} – 4x + 11 = 0

∴ x^{2} – 4x + 11 is a factor of the given equation.

3x + 8 = 0

⇒ x = \(-\frac{8}{3}\)

∴ The roots of the given equation are 2 ± √7i, \(-\frac{8}{3}\)

Question 3.

Solve x^{4} – 4x^{2} + 8x + 35 = 0, given that 2 + i√3 is a root.

Solution:

Let 2 + i√3 be one root

⇒ 2 – i√3 is another root.

The equation having roots 2 ± i√3 is x^{2} – 4x + 7 = 0

∴ x^{2} – 4x + 7 is a factor of x^{4} – 4x^{2} + 8x + 35

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

Question 4.

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

Solution:

2 + √3 is one root

⇒ 2 – √3 is another root.

The equation having the roots of 2 ± √3 is x^{2} – 4x + 1 = 0

∴ x^{2} – 4x + 1 is a factor of x^{4} – 6x^{3} + 11x^{2} – 10x + 2 = 0

∴ The roots of the required equation are 2 ± √3, 1 ± i

Question 5.

Given that -2 + √-7 is a root of the equation x^{4} + 2x^{2} – 16x + 77 = 0, solve it completely.

Solution:

-2 – √-7 (i.e) -2 + i√7 is one root

⇒ -2 – i√7 is another root.

The equation having the roots of -2 ± i√7 is x^{2} + 4x + 11 = 0

∴ x^{2} + 4x + 11 is a factor of x^{4} + 2x^{2} – 16x + 77

∴ The roots of the required equation are -2 ± i√7, 2 ± √3i

Question 6.

Solve the equations x^{4} + 2x^{3} – 16x^{2} – 22x + 7 = 0, given that 2 – √3 is a root of it.

Solution:

2 + √3 is a root ⇒ 2 – √3 is also a root.

The equation having roots 2 ± √3 is

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

⇒ x^{2} – 4x + 1 = 0

x^{2} + 6x + 7 = 0

⇒ x = \(\frac{-6 \pm \sqrt{36-28}}{2}\)

⇒ x = -3 ± √2

∴ Roots are 2 ± √3, -3 ± √2

Question 7.

Solve the equation 3x^{5} – 4x^{4} – 42x^{3} + 56x^{2} + 27x – 36 = 0 given that √2 + √5 is one of its roots.

Solution:

√2 + √5 is a root

⇒ -√2 – √5, -√2 + √5, -√2 – √5 are also roots.

The equation haying roots √2 + √5 is

x^{2} – (√2 + √5 + √2 – √5) + (√2 + √5) (√2 – √5) = 0

⇒ x^{2} – 2√2x – 3 = 0

The equation having roots -√2 ± √5 is

x^{2} – (-√2 + √5 – √2 – √5) + (√2 + √5)(-√2 – √5) = 0

⇒ x^{2} + 2√x – 3 = 0

The equation having roots ±√2 ± √5 is (x^{2} + 2√2x – 3) (x^{2} – 2√2x – 3) = 0

⇒ (x^{2} – 3)^{2} – (2√2x)^{2} = 0

⇒ x^{4} – 6x^{2} + 9 – 8x^{2} = 0

⇒ x^{4} – 14x^{2} + 9 = 0

∴ 3x^{5} – 4x^{4} – 42x^{3} + 56x^{2} + 27x – 36 = 0

⇒ 3x(x^{4} – 14x^{2} + 9) – 4(x^{4} – 14x^{2} + 9) = 0

⇒ (x^{4} – 14x^{2} + 9) (3x – 4) = 0

⇒ x = ±√2 ± √5 or 4/3

∴ The roots are ±√2 ± √5, 4/3

Question 8.

Solve the equation x^{4} – 9x^{3} + 27x^{2} – 29x + 6 = 0, given that one root of it is 2 – √3.

Solution:

2 – √3 is one root ⇒ 2 + √3 is another root.

The equation having the roots of 2 ± √3 is x^{2} – 4x + 1 = 0

∴ x^{2} – 4x + 1 is a factor of the given equation.

x^{2} – 5x + 6 = 0

⇒ (x – 2) (x – 3) = 0

⇒ x = 2, 3

∴ The roots of the required equations are 2 ± √3, 2, 3

Question 9.

Show that the equation \(\frac{a^{2}}{x-a^{\prime}}+\frac{b^{2}}{x-b^{\prime}}+\frac{c^{2}}{x-c^{\prime}}+\ldots+\frac{k^{2}}{x-k^{\prime}}\)= x – m Where a, b, c ….k, m, a’, b’, c’…. k’ are all real numbers, cannot have a non-real root.

Solution:

Let α + iβ be the root of the given equation.

Suppose if β ≠ 0, then α – iβ is also a root of the given equation.

Substitute α + iβ in the given equation, and we get

\(\left[\frac{a^{2}}{\left(\alpha-a^{\prime}\right)^{2}+\beta^{2}}+\frac{b^{2}}{\left(\alpha-b^{\prime}\right)^{2}+\beta^{2}}+\ldots \frac{k^{2}}{\left(\alpha-k^{\prime}\right)^{2}+\beta^{2}}+1\right]\)

= 0

⇒ β = 0

This is a contradiction.

∴ The given equation cannot have non-real roots.