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
⇒ (x2 – 1) (x – 3) = 0
⇒ x3 – 3x2 – 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 – 4i2
= (x – 1)2 + 4
= x2 – 2x + 1 + 4
= x2 – 2x + 5
(x – 4) (x – 2) = x2 – 4x – 2x + 8 = x2 – 6x + 8
Required equation (x2 – 2x + 5) (x2 – 6x + 8) = 0
⇒ x4 – 2x3 + 5x2 – 6x3 + 12x2 – 30x + 8x2 – 16x + 40 = 0
⇒ x4 – 8x3 + 25x2 – 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
= x2 – 4x + 4 – 3
= x2 – 4x + 1
[x – (1 + 2i)] [x – (1 – 2i)] = [(x – 1) – 2i] [(x – 1) + 2i]
= (x – 1)2 – 4i2
= x2 – 2x + 1 + 4
= x2 – 2x + 5
Substituting in (1), the required equation is
(x2 – 4x + 1) (x2 – 2x + 5) = 0
⇒ x4 – 4x3 + x2 – 2x3 + 8x2 – 2x + 5x2 – 20x + 5 = 0
⇒ x4 – 6x3 + 14x2 – 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
⇒ x2 (x – 2)2 (x + 2)2 = 0
⇒ x2 (x2 – 4)2 = 0
⇒ x2 (x4 – 8x2 + 16) = 0
⇒ x6 – 8x4 + 16x2 = 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
= x2 – 2x + 1 – 3
= x2 – 2x – 2
(x – 2) (x – 5) = x2 – 2x – 5x + 10 = x2 – 7x + 10
Substituting in (1), the required equation is
(x2 – 2x – 2) (x2 – 7x + 10) = 0
⇒ x4 – 2x3 – 2x2 – 7x3 + 14x2 + 14x + 10x2 – 20x – 20 = 0
⇒ x4 – 9x3 + 22x2 – 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 4x3 – 6x2 + 7x + 3 = 0, then find the value of αβ + βγ + γα.
Solution:
α, β, γ are the roots of 4x3 – 6x2 + 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 x3 – 6x2 + 9x – 4 = 0, then find α.
Solution:
1, 1, α are roots of x3 – 6x2 + 9x – 4 = 0
Sum = 1 + 1 + α = 6
⇒ α = 6 – 2 = 4

Question 4.
If -1, 2 and α are the roots of 2x3 + x2 – 7x – 6 = 0, then find α.
Solution:
-1, 2, α are roots of 2x3 + x2 – 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 x3 – 2x2 + ax + 6 = 0, then find a.
Solution:
1, -2 and 3 are roots of x3 – 2x2 + ax + 6 = 0
⇒ 1(-2) + (-2)3 + 3 . 1 = a
⇒ a = -2 – 6 + 3 = -5

Question 6.
If the product of the roots of 4x3 + 16x2 – 9x – a = 0 is 9, then find a.
Solution:
α, β, γ are the roots of 4x3 + 16x2 – 9x – a = 0
αβγ = 9
⇒ $$\frac{a}{4}$$ = 9
⇒ a = 36

Question 7.
Find the values of s1, s2, s3, and s4 for each of the following equations.
(i) x4 – 16x3 + 86x2 – 176x + 105 = 0
(ii) 8x4 – 2x3 – 27x2 + 6x + 9 = 0

Solution:
(i) Given equation is x4 – 16x3 + 86x2 – 176x + 105 = 0
We know that

(ii) Equation is 8x4 – 2x3 – 27x2 + 6x + 9 = 0

II.

Question 1.
If α, β and 1 are the roots of x3 – 2x2 – 5x + 6 = 0, then find α and β.
Solution:
α, β and 1 are the roots of x3 – 2x2 – 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
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 x3 – 2x2 + 3x – 4 = 0, then find
(i) Σα2β2
(ii) Σαβ(α + β)
Solution:
Since α, β, γ are the roots of x3 – 2x2 + 3x – 4 = 0 then
α + β + γ = 2
αβ + βγ + γα = 3
αβγ = 4
(i) Σα2β2 = α2β2 + β2γ2 + γ2α2
= (αβ + βγ + γα)2 – 2αβγ(α + β + γ)
= 9 – 2 . 2 . 4
= 9 – 16
= -7

(ii) Σαβ(α + β) = α2β + β2γ + γ2α + αβ2 + βγ2 + γα2
= (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 x3 + px2 + 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) Σα3β3
Solution:
α, β and γ are the roots of x3 + px2 + 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)
= -(p3 + 4p2 (α + β + γ) + 16p (αβ + βγ + γα) + (64αβγ)
= -(p3 – 4p3 + 16pq – 64r)
= 3p3 – 16pq + 64r

(iv) Σα3β3 = α3β3 + β3γ3 + γ3α3
(αβ + βγ + γα)2 = α2β2 + β2γ2 + γ2α2 + 2αβγ (α + β + γ)
⇒ q2 = α2β2 + β2γ2 + γ2α2 + 2pr
⇒ α2β2 + β2γ2 + γ2α2 = q2 – 2pr
∴ α3β3 + β3γ3 + γ3α3 = (α2β2 + β2γ2 + γ2α2) (αβ + βγ + γα) – αβγ Σα2β
= (q2 – 2pr) . q + r[(αβ + βγ + γα) (α + β + γ) – 3αβγ]
= q3 – 2pqr + r(-pq + 3r)
= q3 – 2pqr – pqr + 3r2
= q3 – 3pqr + 3r2

III.

Question 1.
If α, β, γ are the roots of x3 – 6x2 + 11x – 6 = 0, then find the equation whose roots are α2 + β2, β2 + γ2, γ2 + α2.
Solution:
1st Method:
Let α, β, γ are the roots of the equation x3 – 6x2 + 11x – 6 = 0
∴ α + β + γ = 6, αβ + βγ + γα = 11
Let y = α2 + β2 = α2 + β2 + γ2 – γ2
⇒ y = (α + β + γ)2 – 2(αβ + βγ + γα) – x2
⇒ y = 36 – 22 – x2
⇒ x2 = 14 – y
⇒ x = $$\sqrt{14-y}$$
Substitute x = $$\sqrt{14-y}$$ in x3 – 6x2 + 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 + y2) = (14 – y)(y2 – 50y + 625)
⇒ 8100 – 1080y + 36y2 = 14y2 – 700y + 8750 – y3 + 50y2 – 625y
⇒ 8100 – 1080y + 36y2 = -y3 + 64y2 – 1325y + 8750
⇒ y3 – 28y2 + 245y – 650 = 0
∴ The required equation is x3 – 28x2 + 245x – 650 = 0
2nd Method:
Let α, β, γ are the roots of x3 – 6x2 + 11x – 6 = 0
It is an odd-degree reciprocal equation of class two.
∴ x – 1 is a factor of x3 – 6x2 + 11x – 6

∴ x3 – 6x2 + 11x – 6 = (x – 1) (x2 – 5x + 6) = (x – 1) ( x – 2) (x – 3)
∴ The roots of x3 – 6x2 + 11x – 6 = 0 are α = 1, β = 2, γ = 3
Now α2 + β2 = 12 + 22 = 5
β2 + γ2 = 22 + 32 = 13
γ2 + α2 = 32 + 12 = 10
Therefore the cubic equation with roots α2 + β2, β2 + γ2, γ2 + α2 is (x – 5) (x – 13) (x – 10) = 0
⇒ x3 – (5 + 13 + 10) x2 + (65 + 130 + 50)x – 650 = 0
⇒ x3 – 28x2 + 245x – 650 = 0

Question 2.
If α, β, γ are the roots of x3 – 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 x3 – 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 = γ2 + $$\frac{24}{\gamma}$$
⇒ y = x2 + $$\frac{24}{x}$$
⇒ xy = x3 + 24
⇒ xy = 7x – 6 + 24 [from (1)]
⇒ x(y – 7) = 18
⇒ x = $$\frac{18}{y-7}$$
Substituting x = $$\frac{18}{y-7}$$ in x3 – 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(y2 – 14y + 49) + 6(y3 – 21y2 + 147y – 343) = 0
⇒ 972 – 21(y2 – 14y + 49) + (y3 – 21y2 + 147y – 343) = 0
⇒ y3 – 42y2 + 441y – 400 = 0
∴ The equation with roots (α – β)2, (β – γ)2, (γ – α)2 is x3 – 42x2 + 441x – 400 = 0
2nd Method:
α, β, γ are the roots of x3 – 7x + 6 = 0
By trial and error method x = 1 satisfies this equation.
∴ x – 1 is a factor of x3 – 7x + 6

∴ x3 – 7x + 6 = (x – 1) (x2 + x – 6) = (x – 1)(x + 3)(x – 2)
∵ α, β, γ are the roots of x3 – 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
⇒ x3 – (16 + 25 + 1) x2 + (400 + 25 + 16)x – 400 = 0
⇒ x3 – 42x2 + 441x – 400 = 0

Question 3.
If α, β, γ are the roots of x3 – 3ax + b = 0, prove that Σ(α – β) (α – γ) = 9a.
Solution:
α, β, γ are the roots of x3 – 3ax + b = 0
∴ α + β + γ = 0, αβ + βγ + γα = -3a, αβγ = -b
Σ(α – β) (α – γ) = Σ[α2 – α(β + γ) + βγ]
= Σ[α2 + α2 + βγl
= 2(α2 + β2 + γ2) + (βγ + γα + αβ)
= 2(α + β + γ)2 – 4(αβ + βγ + γα) + (αβ + βγ + γα)
= 0 – 4(-3a) + (-3a)
= 9a