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

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

I.

Question 1.

Form polynomial equations of the lowest degree, with roots as given below.

(i) 1, -1, 3

Solution:

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

Sol. Required equation is (x – 1) (x + 1) (x – 3) = 0

⇒ (x^{2} – 1) (x – 3) = 0

⇒ x^{3} – 3x^{2} – x + 3 = 0

(ii) 1 ± 2i, 4, 2

Solution:

In an equation, imaginary roots occur in conjugate pairs.

Equation having roots α, β, γ, δ is (x – α) (x – β) (x – γ) (x – δ) = 0

Required equation is [x – (1 + 2i)] [x – (1 – 2i)] (x – 4) (x – 2) = 0

(x – (1 + 2i)] [x – (1 – 2i)] = [(x – 1) – 2i] [(x – 1) + 2i]

= (x – 1)^{2} – 4i^{2}

= (x – 1)^{2} + 4

= x^{2} – 2x + 1 + 4

= x^{2} – 2x + 5

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

Required equation (x^{2} – 2x + 5) (x^{2} – 6x + 8) = 0

⇒ x^{4} – 2x^{3} + 5x^{2} – 6x^{3} + 12x^{2} – 30x + 8x^{2} – 16x + 40 = 0

⇒ x^{4} – 8x^{3} + 25x^{2} – 46x + 40 = 0

(iii) 2 ± √3, 1 ± 2i

Solution:

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

[x – (2 + √3)] [x – (2 – √3)]

= [(x – 2) – √3] [(x – 2) + √3]

= (x – 2)^{2} – 3

= x^{2} – 4x + 4 – 3

= x^{2} – 4x + 1

[x – (1 + 2i)] [x – (1 – 2i)] = [(x – 1) – 2i] [(x – 1) + 2i]

= (x – 1)^{2} – 4i^{2}

= x^{2} – 2x + 1 + 4

= x^{2} – 2x + 5

Substituting in (1), the required equation is

(x^{2} – 4x + 1) (x^{2} – 2x + 5) = 0

⇒ x^{4} – 4x^{3} + x^{2} – 2x^{3} + 8x^{2} – 2x + 5x^{2} – 20x + 5 = 0

⇒ x^{4} – 6x^{3} + 14x^{2} – 22x + 5 = 0

(iv) 0, 0, 2, 2, -2, -2

Solution:

Required equation is (x – 0) (x – 0) (x – 2) (x – 2) (x + 2) (x + 2) = 0

⇒ x^{2} (x – 2)^{2} (x + 2)^{2} = 0

⇒ x^{2} (x^{2} – 4)^{2} = 0

⇒ x^{2} (x^{4} – 8x^{2} + 16) = 0

⇒ x^{6} – 8x^{4} + 16x^{2} = 0

(v) 1 ± √3, 2, 5

Solution:

Required equation is [x – (1 + √3)] [x – (1 – √3)][(x – 2) (x – 5)] = 0 ………(1)

[x – (1 + √3)] [x – (1 – √3)] = [(x – 1) – √3] [(x – 1) + √3]

= (x – 1)^{2} – 3

= x^{2} – 2x + 1 – 3

= x^{2} – 2x – 2

(x – 2) (x – 5) = x^{2} – 2x – 5x + 10 = x^{2} – 7x + 10

Substituting in (1), the required equation is

(x^{2} – 2x – 2) (x^{2} – 7x + 10) = 0

⇒ x^{4} – 2x^{3} – 2x^{2} – 7x^{3} + 14x^{2} + 14x + 10x^{2} – 20x – 20 = 0

⇒ x^{4} – 9x^{3} + 22x^{2} – 6x – 20 = 0

(vi) 0, 1, \(-\frac{3}{2}\), \(-\frac{5}{2}\)

Solution:

Required equation is

Question 2.

If α, β, γ are the roots of 4x^{3} – 6x^{2} + 7x + 3 = 0, then find the value of αβ + βγ + γα.

Solution:

α, β, γ are the roots of 4x^{3} – 6x^{2} + 7x + 3 = 0

α + β + γ = \(-\frac{a_{1}}{a_{0}}=\frac{6}{4}\)

αβ + βγ + γα = \(\frac{a_{2}}{a_{0}}=\frac{7}{4}\)

αβγ = \(-\frac{a_{3}}{a_{0}}=-\frac{3}{4}\)

∴ αβ + βγ + γα = \(\frac{7}{4}\)

Question 3.

If 1, 1, α are the roots of x^{3} – 6x^{2} + 9x – 4 = 0, then find α.

Solution:

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

Sum = 1 + 1 + α = 6

⇒ α = 6 – 2 = 4

Question 4.

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

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 5.

If 1, -2 and 3 are roots of x^{3} – 2x^{2} + ax + 6 = 0, then find a.

Solution:

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

⇒ 1(-2) + (-2)3 + 3 . 1 = a

⇒ a = -2 – 6 + 3 = -5

Question 6.

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

Solution:

α, β, γ are the roots of 4x^{3} + 16x^{2} – 9x – a = 0

αβγ = 9

⇒ \(\frac{a}{4}\) = 9

⇒ a = 36

Question 7.

Find the values of s_{1}, s_{2}, s_{3}, and s_{4} for each of the following equations.

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

(ii) 8x^{4} – 2x^{3} – 27x^{2} + 6x + 9 = 0

Solution:

(i) Given equation is x^{4} – 16x^{3} + 86x^{2} – 176x + 105 = 0

We know that

(ii) Equation is 8x^{4} – 2x^{3} – 27x^{2} + 6x + 9 = 0

II.

Question 1.

If α, β and 1 are the roots of x^{3} – 2x^{2} – 5x + 6 = 0, then find α and β.

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 + β = 1

⇒ β = 1 – α

= 1 – 3

= -2

∴ α = 3 and β = -2

Question 2.

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 3.

If α, β and γ are the roots of x^{3} + px^{2} + qx + r = 0, then find the following.

(i) \(\sum \frac{1}{\alpha^{2} \beta^{2}}\)

(ii) \(\frac{\beta^{2}+\gamma^{2}}{\beta \gamma}+\frac{\gamma^{2}+\alpha^{2}}{\gamma \alpha}+\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}\) or \(\Sigma \frac{\beta^{2}+\gamma^{2}}{\beta \gamma}\)

(iii) (β + γ – 3α) (γ + α – 3β) (α + β – 3γ)

(iv) Σα^{3}β^{3}

Solution:

α, β and γ are the roots of x^{3} + px^{2} + qx + r = 0,

α + β + γ = -p

αβ + βγ + γα = q

αβγ = -r

(iii) (β + γ – 3α) (γ + α – 3β) (α + β – 3γ) = (α + β + γ – 4α) (α + β + γ – 4β) (α + β + γ – 4γ)

= (-p – 4α) (-p – 4β) (-p – 4γ)

= -(p + 4α) (p + 4β) (p + 4γ)

= -(p^{3} + 4p^{2} (α + β + γ) + 16p (αβ + βγ + γα) + (64αβγ)

= -(p^{3} – 4p^{3} + 16pq – 64r)

= 3p^{3} – 16pq + 64r

(iv) Σα^{3}β^{3} = α^{3}β^{3} + β^{3}γ^{3} + γ^{3}α^{3}

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

⇒ q^{2} = α^{2}β^{2} + β^{2}γ^{2} + γ^{2}α^{2} + 2pr

⇒ α^{2}β^{2} + β^{2}γ^{2} + γ^{2}α^{2} = q^{2} – 2pr

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

= (q^{2} – 2pr) . q + r[(αβ + βγ + γα) (α + β + γ) – 3αβγ]

= q^{3} – 2pqr + r(-pq + 3r)

= q^{3} – 2pqr – pqr + 3r^{2}

= q^{3} – 3pqr + 3r^{2}

III.

Question 1.

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:

1st Method:

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

∴ α + β + γ = 6, αβ + βγ + γα = 11

Let y = α^{2} + β^{2} = α^{2} + β^{2} + γ^{2} – γ^{2}

⇒ y = (α + β + γ)^{2} – 2(αβ + βγ + γα) – x^{2}

⇒ y = 36 – 22 – x^{2}

⇒ x^{2} = 14 – y

⇒ x = \(\sqrt{14-y}\)

Substitute x = \(\sqrt{14-y}\) in x^{3} – 6x^{2} + 11x – 6 = 0

⇒ (\(\sqrt{14-y}\))3 – 6(\(\sqrt{14-y}\))2 + 11(\(\sqrt{14-y}\)) – 6 = 0

⇒ (14 – y) \(\sqrt{14-y}\) – 6(14 – y) + 11 \(\sqrt{14-y}\) – 6 = 0

⇒ -6(14 – y + 1) = \(\sqrt{14-y}\) [-11 – 14 + y]

⇒ -6(15 – y) = (\(\sqrt{14-y}\)) (y – 25)

Squaring on both sides

i.e., [-6(15 – y)]^{2} = [\(\sqrt{14-y}\)(y – 25)]^{2}

⇒ 36(225 – 30y + y^{2}) = (14 – y)(y^{2} – 50y + 625)

⇒ 8100 – 1080y + 36y^{2} = 14y^{2} – 700y + 8750 – y^{3} + 50y^{2} – 625y

⇒ 8100 – 1080y + 36y^{2} = -y^{3} + 64y^{2} – 1325y + 8750

⇒ y^{3} – 28y^{2} + 245y – 650 = 0

∴ The required equation is x^{3} – 28x^{2} + 245x – 650 = 0

2nd Method:

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

It is an odd-degree reciprocal equation of class two.

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

∴ x^{3} – 6x^{2} + 11x – 6 = (x – 1) (x^{2} – 5x + 6) = (x – 1) ( x – 2) (x – 3)

∴ The roots of x^{3} – 6x^{2} + 11x – 6 = 0 are α = 1, β = 2, γ = 3

Now α^{2} + β^{2} = 1^{2} + 2^{2} = 5

β^{2} + γ^{2} = 2^{2} + 3^{2} = 13

γ^{2} + α^{2} = 3^{2} + 1^{2} = 10

Therefore the cubic equation with roots α^{2} + β^{2}, β^{2} + γ^{2}, γ^{2} + α^{2} is (x – 5) (x – 13) (x – 10) = 0

⇒ x^{3} – (5 + 13 + 10) x^{2} + (65 + 130 + 50)x – 650 = 0

⇒ x^{3} – 28x^{2} + 245x – 650 = 0

Question 2.

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:

1st Method:

Let α, β, γ are the roots of the equation x^{3} – 7x + 6 = 0 …….(1)

α + β + γ = 0, αβγ = -6

Let y = (α – β)^{2} = (α + β)^{2} – 4αβ

⇒ y = (-γ)^{2} – 4(\(\frac{6}{\gamma}\))

⇒ y = γ^{2} + \(\frac{24}{\gamma}\)

⇒ y = x^{2} + \(\frac{24}{x}\)

⇒ xy = x^{3} + 24

⇒ xy = 7x – 6 + 24 [from (1)]

⇒ x(y – 7) = 18

⇒ x = \(\frac{18}{y-7}\)

Substituting x = \(\frac{18}{y-7}\) in x^{3} – 7x + 6 = 0

(\(\frac{18}{y-7}\))3 – 6(\(\frac{18}{y-7}\)) + 6 = 0

⇒ (18)^{3} – 7(18) (y – 7)^{2} + 6(y – 7)^{3} = 0

⇒ 5832 – 126(y^{2} – 14y + 49) + 6(y^{3} – 21y^{2} + 147y – 343) = 0

⇒ 972 – 21(y^{2} – 14y + 49) + (y^{3} – 21y^{2} + 147y – 343) = 0

⇒ y^{3} – 42y^{2} + 441y – 400 = 0

∴ The equation with roots (α – β)^{2}, (β – γ)^{2}, (γ – α)^{2} is x^{3} – 42x^{2} + 441x – 400 = 0

2nd Method:

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

By trial and error method x = 1 satisfies this equation.

∴ x – 1 is a factor of x^{3} – 7x + 6

∴ x^{3} – 7x + 6 = (x – 1) (x^{2} + x – 6) = (x – 1)(x + 3)(x – 2)

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

α = 1, β = -3, γ = 2,

Now (α – β)^{2} = [1 – (-3)]^{2} = (4)^{2} = 16

(β – γ)^{2} = [-3 – 2]^{2} = 25

(γ – α)^{2} = [2 – 1]^{2} = 1

∴ The cubic equation whose roots are (α – β)^{2}, (β – γ)^{2}, (γ – α)^{2} is (x – 16) (x – 25) (x – 1) = 0

⇒ x^{3} – (16 + 25 + 1) x^{2} + (400 + 25 + 16)x – 400 = 0

⇒ x^{3} – 42x^{2} + 441x – 400 = 0

Question 3.

If α, β, γ are the roots of x^{3} – 3ax + b = 0, prove that Σ(α – β) (α – γ) = 9a.

Solution:

α, β, γ are the roots of x^{3} – 3ax + b = 0

∴ α + β + γ = 0, αβ + βγ + γα = -3a, αβγ = -b

Σ(α – β) (α – γ) = Σ[α^{2} – α(β + γ) + βγ]

= Σ[α^{2} + α^{2} + βγl

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

= 2(α + β + γ)^{2} – 4(αβ + βγ + γα) + (αβ + βγ + γα)

= 0 – 4(-3a) + (-3a)

= 9a