TS Inter Second Year Maths 2A Theory of Equations Important Questions Long Answer Type

Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Theory of Equations Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2A Theory of Equations Important Questions Long Answer Type

Question 1.
If the roots of the equation x³ + 3px² + 3qx + r = 0 are in AP then show that 2p³ – 3pq + r = 0. [May ’08]
Solution:
Given equation is x³ + 3px² + 3qx + r = 0
Since the roots are in A.P then they must be of the form a – d, a, a + d.
Now, s₁ = a – d + a + a + d = \(\frac{-3 p}{1}\)
3a = – 3p
a = – p
Since a is a root of x³ + 3px² + 3qx + r = 0 then
(- p)³ + 3p (- p)² + 3q(- p) + r = 0
⇒ – p³ + 3p³ – 3pq + r = 0
⇒ 2p³ – 3pq + r = 0.

Question 2.
If the roots of the equation x³ + 3px² + 3qx + r = 0 are in G.P then show that p³r = q³. [March ‘03]
Solution:
Given equation is x³ + 3px² + 3qx + r = 0
Since the roots are in G.P then they must be of the form \(\frac{a}{r}\), a, ar
Now, s₃ = \(\frac{a}{r}\) . a . ar
= \(\frac{-\mathrm{r}}{1}\) = – r
⇒ a³ = – r
⇒ a = (- r)^1/3
Since a is a root of x³ + 3px² + 3qx + r = 0 then
\(\left((-r)^{1 / 3}\right)^3+3 p\left((-r)^{1 / 3}\right)^2+3 q\left((-r)^{1 / 3}\right)\) + r = 0
⇒ – r + 3p (- r)^2/3 + 3q(- r)^1/3 + r = 0
⇒ 3p(- r)^2/3 = – 3q(- r)^1/3
Cubing on both sides,
⇒ p³ (- r)² = – q³(- r)
⇒ p³ r² = q³r
⇒ p³r = q³

Question 3.
If the roots of the equation x³ + 3px² + 3q + r = 0 are in H.P then show that 2q³ = r (3pq – r).
Solution:
Given equation is x³ + 3px² + 3qx + r = 0
The roots of x³ + 3px² + 3qx + r = 0 are in H.P.
The roots of \(\left(\frac{1}{x}\right)^3+3 p\left(\frac{1}{x}\right)^2+3 q\left(\frac{1}{x}\right)\) + r = 0 are in A.p.
\(\frac{1}{x^3}+\frac{3 p}{x^2}+\frac{3 q}{x}\) + r = 0 are in A.P
⇒ 1 + 3px + 3qx² + rx³ = o
⇒ rx³ + 3qx² + 3px + 1 = 0 ……………..(1)
Let the roots of (1) be a – d, a, a + d
s₁ = a – d + a + a + d = \(\frac{-3 q}{r}\)
⇒ 3a = \(\frac{-3 q}{r}\)
a = \(\frac{-q}{r}\)
Since a is a root of (1)
⇒ \(r \cdot \frac{-q^3}{r^3}+3 q \cdot \frac{q^2}{r^2}+3 p \cdot \frac{-q}{r}+1\) = 0
⇒ \(\frac{-q^3}{r^2}+\frac{3 q^3}{r^2}-\frac{3 p q}{r}\) + 1 = 0
⇒ 2q³ = r (3pq – r). +
Hence proved.

Question 4.
Solve 4x³ – 24x² + 23x + 18 = 0, given that the roots of this equation are in arithmetic
progression. [Mar. ’14, May ’06, ’01]
Solution:
Given equation is 4x³ – 24x² + 23x + 18 = 0
Given that the roots of this equation are in A.P.
Since the roots are in A.P, they must be of the form, a – d, a, a + d.
Now, a – d + a + a + d = \(-\frac{-24}{4}\) = 6
⇒ 3a = 6
⇒ a = 2
(a – d) a (a + d) = \(\frac{-18}{4}\)
(a² – d²)a = \(\frac{-9}{2}\)
(4 – d²)2 = \(\frac{-9}{2}\)
4 – d² = \(\frac{-9}{4}\)
d² = 4 + \(\frac{9}{4}\) = \(\frac{25}{4}\)
d = ± \(\frac{5}{2}\)

Case – I:
If a = 2, d = \(\frac{5}{2}\)
The roots of given equation are – \(\frac{-1}{2}\), 2, \(\frac{9}{2}\).

Case-2:
If a = 2, d = \(\frac{-5}{2}\)
The roots of given equation are \(\frac{9}{2}\), 2, \(\frac{-1}{2}\).

Question 5.
Solve x³ – 3x² – 6x + 8 = 0, given that the : roots are in A.P. [March ’07]
Solution:
Given equation is x³ – 3x² – 6x + 8 = 0
Since, the roots are in A.P., thus they must; be of the form a – d, a, a + d.
Now, a – d + a + a + d = \(\frac{-(-3)}{1}\) = 3
⇒ 3a = 3
⇒ a = 1
⇒ (a – d) a (a + d) = \(\frac{-8}{1}\)
a² – d² = – 8
1 – d² = – 8
d = 9
⇒ d = ±3
Case – 1 : If d = 3; a = 1
The roots of the given equation are; a – d, a, a + d = – 2, 1, 4.

Case – 2: If d = – 3; – a = 1;
The roots of the given equation are (a – d) a, (a + d) = 4, 1, – 2 .

Question 6.
Solve x³ – 7x² + 14x – 8 = 0, given that the roots are in geometric progression. [May ’07]
Solution:
Given equation is x³ – 7x² + 14x – 8 = 0
Given that the roots are in G.P then they may be of the form \(\frac{a}{r}\), a, ar
\(\frac{a}{r}\) + a + ar = a [\(\frac{1}{r}\) + 1 + r] = \(\frac{-(-7)}{1}\)
⇒ a[\(\frac{1}{r}\) + 1 + r] = 7 …………..(1)
\(\frac{a}{r}\) . a . ar = 1
⇒ a³ = 8
⇒ a = 2
(1) ⇒ 2\(\left(\frac{r^2+r+1}{r}\right)\) = 7
⇒ 2r² – 5r + 2 = 0
⇒ 2r² – 4r – r + 2 = 0
⇒ 2r (r – 2) – 1 (r – 2) = 0
⇒ (2r – 1) (r – 2) = 0
⇒ r = \(\frac{1}{2}\) (or) r = 2
Case – 1:
If a = 2, r = \(\frac{1}{2}\)
The roots of given equation are \(\frac{a}{r}\), a, ar = 4, 2, 1
Case – 2 :
If a = 2, r = 2
The roots of given equation are \(\frac{a}{r}\), a, ar = 1, 2, 4.

Question 7.
Solve the equation 15x³ – 23x² + 9x – 1 = 0, the roots being in H.P. [May ’02]
Solution:
Giveti equation is 15x³ – 23x² + 9x – 1 = 0
The roots of 15x³ – 23x² + 9x – 1 = 0 are in H.P
The roots of
\(15 \cdot \frac{1}{x^3}-23 \cdot \frac{1}{x^2}+9 \cdot \frac{1}{x}-1\) = 0 are in A.P.
15 – 23x + 9x² – x³ = 0
⇒ x³ – 9x² + 23x – 15 = 0
Let the roots be a – d, a, a + d
a – d + a + a + d = \(\frac{-(-9)}{1}\) = 9
⇒ 3a = 9
⇒ a = 3
(a – d) a (a + d) = \(\frac{-(-15)}{1}\) = 15
⇒ a(a² – d²) = 15
⇒ 3(9 – d²) = 15
9 – d² = 5
d² = 4
d = 21
The roots of x³ – 9x² + 23x – 15 = 0 are a – d, a, a + d =3 – 2, 3, 3 + 2 = 1, 3, 5
The roots of 15x³ – 23x² + 9x – 1 = 0 are \(\frac{1}{1}, \frac{1}{3}, \frac{1}{5}\) = 1, \(\frac{1}{3}, \frac{1}{5}\).

Question 8.
Solve 18x³ + 81x² + 121x + 60 = 0 given that one root is equal to half the sum of the remaining roots. [May ’11, March ‘05] [AP – Mar.2019]
Solution:
Given that one root is equal to half of the sum of the remaining roots.
Given equation is 18x³ + 81x² + 121x + 60 = 0
i.e., the roots are in A.P.
Let a – d, a, a + d be the roots of the given equation.
∴ a – d + a + a + d = \(\frac{-81}{18}\)
3a = \(\frac{-81}{18}\)
a = \(\frac{-3}{2}\)
(a – d) a (a + d) = \(\frac{-60}{18}\)
a(a – d) = \(\frac{-81}{18}\)
\(\frac{-3}{2}\left(\frac{9}{4}-d^2\right)=\frac{-60}{18}\)
d² = \(\frac{9}{4}-\frac{20}{9}=\frac{81-80}{36}=\frac{1}{36}\)
d = \(\frac{1}{6}\)
∴ The roots of the given equation are a – d, a, a + d = \(\frac{-3}{2}-\frac{1}{6}, \frac{-3}{2}, \frac{-3}{2}+\frac{1}{6}\)
= \(\frac{-5}{3}, \frac{-3}{2}, \frac{-4}{3}\)

Question 9.
Given that one root of 2x³ + 3x² – 8x + 3 = 0 is double the other root, find the roots of the equation. [May ’03]
Solution:
Given equation is 2x³ + 3x² – 8x + 3 = 0
Now, α + β + γ = \(\frac{-3}{2}\) …………..(1)
αβ + βγ + γα = \(\frac{-8}{2}\) = – 4 ……………(2)
αβγ = \(\frac{-3}{2}\) ……………..(3)
Given one root is double the other then
α = 2β
(1) ⇒ 2β + β + γ = \(\frac{-3}{2}\)
⇒ 3β + γ = \(\frac{-3}{2}\)
⇒ γ = \(\frac{-3}{2}\) – 3β …………….(4)
(2) ⇒ 2β . β + βγ + γ . 2β = – 4
2β² + βγ + 2βγ = – 4
2β² + 3βγ = – 4 ……………..(5)
From (4) & (5)
2β² + 3β(\(\frac{-3}{2}\) – 3β) = – 4
2β² – \(\frac{9 \beta}{2}\) – 9β²=
⇒ \(\frac{4 \beta^2-9 \beta-18 \beta^2}{2}\) = – 4
⇒ – 14β² – 9β = – 8
⇒ 14β² + 9β – 8 = 0
⇒ 14β² + 16β – 7β – 8 = 0
⇒ (2β – 1) (7β + 8) = 0
⇒ β = \(\frac{-8}{7}\) (or) β = \(\frac{1}{2}\)
β = \(\frac{-8}{7}\) does not satisfy the given equation.
β = \(\frac{1}{2}\)
⇒ α = 2 . \(\frac{1}{2}\) = 1
Substitute the value of β in equation (4)
⇒ γ = \(\frac{-3}{2}\) – 3 . \(\frac{1}{2}\)
= \(\frac{-3-3}{2}=\frac{-6}{2}\) = – 3.
∴ The roots of the given equation are 1, \(\frac{1}{2}\), – 3.

Question 10.
Solve x⁴ + x³ – 16x² – 4x + 48 = 0, given that the product of two of the roots is 6. [TS – Mar. 2019; May ’12. ’09]
Solution:
Given equation is x⁴ + x³ – 16x² – 4x + 48 = 0
Let α, β, γ, δ are the roots of the given equation.
Then αβγδ = \(\frac{48}{1}\) = 48
Given that the product of the two roots of the given equation is 6.
Suppose that αβ = 6
Now, 6γδ = 48
γδ = 8
Let α + β = p, γ + δ = q
The quadratic equation whose roots are α, β is
x² – x(α + β) + αβ = 0
x² – px + 6 = 0 ………….(1)
The quadratic equation whose roots are γ, δ is
x² – x (γ + δ) + γδ = 0
⇒ x² – qx + 8 = 0 ……………(2)
∴ x⁴ + x³ – 16x² – 4x + 48
= (x² – px + 6) (x² – qx + 8)
= x⁴ – qx³ + 8x² – px³ + pqx² – 8px + 6x² – 6qx + 48
= x⁴ + (- p – q)x³ + (pq+ 14)x² + (- 8p – 6q) x + 48
Now comparing coefficients of x³ on both sides, we get
⇒ – p – q = 1
⇒ p + q = – 1 ………………(3)
Now comparing coefficients of x on both sides, we get
– 8p – 6q = – 4
⇒ 4p + 3q = 2 ……………….(4)
Solve (3) and (4);

⇒ p = \(\frac{-5}{-1}\)
⇒ q = \(\frac{6}{-1}\)
⇒ p = 5, q = – 6
Substitute ‘p’ value in (1),
x² – 5x + 6 = 0
⇒ x² – 3x – 2x + 6 = 0
⇒ x (x – 3) – 2 (x – 3) = 0
⇒ (x – 2) (x – 3) = 0
⇒ x = 2, x = 3
Substitute ‘q’ value in (2),
(2) ⇒ x² + 6x + 8 = 0
⇒ x² + 4x . 2x + 8 = 0
⇒ (x + 2) (x +4) = 0
⇒ x = – 4, x = – 2
∴ The roots of given equation are 3, 2, – 4, – 2.

Question 11.
Given that two roots of 4×3+20×2-23x+6=O are equal, find all the roots of the given equation.
Solution:
Given equation is 4x³ + 20x² – 23x + 6 = 0
Let α, β, γ be the roots of given equation.
Then s₁ = α + β + γ = \(\frac{-(20)}{4}\)
⇒ α + β + γ = – 5 …………….(1)
⇒ αβ + βγ + γα = \(\frac{-23}{4}\) ……………(2)
⇒ αβγ = \(\frac{-6}{4}=\frac{-3}{2}\) ………….(3)
Given that two roots are equal
⇒ β = α
(1) ⇒ 2α + γ = – 5
⇒ γ = – 5 – 2α
(2) ⇒ α² + αy + αy = \(\frac{-23}{4}\)
⇒ α² + 2αγ = \(\frac{-23}{4}\)
⇒ α² + α (- 5 – 2α) = \(\frac{-23}{4}\)
⇒ 4 (α² – 10α – 4α²) = – 23
⇒ 4 (- 3α² – 10α) = – 23
⇒ 12α² + 40α – 23 = 0
⇒ 12α² + 46α – 6α – 23 = 0
⇒ 2α (6α + 23) – 1(6α + 23) = 0
⇒ (6α + 23) (2α – 1) = 0
∴ α = \(\frac{-23}{6}\) does not satisfy given equation.
∴ α = \(\frac{1}{2}\)
⇒ β = \(\frac{1}{2}\)
γ = – 5 – 2 . \(\frac{1}{2}\) = – 6
γ = – 6
∴ The roots of the given equation are \(\frac{1}{2}\), \(\frac{1}{2}\), – 6.

Question 12.
Solve x⁴ + 4x³ – 2x² – 12x + 9 = 0, given that it has two pairs of equal roots. [March ’11, ’03]
Solution:
Given equation is x⁴ + 4x³ – 2x² – 12x + 9 = 0
Given that it has two pairs of equal roots then let the roots of the given equation be
α + α + β + β = \(\frac{-4}{1}\)
⇒ 2 (α + β) = – 4
⇒ α + β = – 2
αα + αβ + βα + βα + αβ + β² = \(\frac{-2}{1}\)
α² + 4αβ + β²
(α + β)² + 2αβ = – 2
4 + 2αβ = – 2
2αβ = – 6
αβ = – 3
We know that,
(α – β)² = (α + β)² – 4αβ
= 4 – 4 (- 3)
= 4 + 12 = 16

α – β = 4
α + β = – 2
⇒ 2α = 2
⇒ α = 1

α + β = – 2
1 + β = – 2
β = – 3
∴ The roots of the given equation are 1, 1, – 3, – 3.

Question 13.
Find the roots of x⁴ – 16x³ + 86x² – 176x + 105 = 0. [March ’08, ’02]
Solution:
Let f(x) = x⁴ – 16x³ + 86x² – 176x + 105 = 0
If x = 1 then f(1)= 1 – 16 + 86 – 176 + 105 = 0
= 192 – 192 = 0
Hence, 1 is a root of f(x) = 0
If x = 3 then
f(3) = 81 – 16 (27) + 86(9) – 176(3) + 105
= 81 – 432+ 774 – 528 + 105
= 960 – 960 = 0
Hence 3 is a root of f(x) = 0.
By the method of synthetic division we have

∴ The quotient is x² – 12x + 35 and remainder is zero.
∴ f(x) = x⁴ – 16x³ + 86x² – 176x + 105
= (x – 1)(x – 3) (x² – 12x + 35)
= (x – 1) (x – 3) (x – 5) (x – 7)
Hence the roots of the given equation are 1, 3, 5, 7.

Question 14.
Solve x⁴ – 10x³ + 26x² – 10x + 1 = 0. [TS – Mar. ‘18; Mar. ‘12, ’10, Board Paper]
Solution:
Given equation is x – 10x³ + 26x² – 10x + 1 = O
This is standard form of reciprocal equation.
The given equation is a reciprocal degree of class – 1 on dividing with x²
⇒ x² – 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
If x + \(\frac{1}{x}\) = a
⇒ x² + \(\frac{1}{x^2}\) = a² – 2
⇒ (a² – 2) – 10a + 26 = 0
⇒ a² – 10a + 24 = 0
⇒ a² – 6a – 4a + 24 = 0
⇒ a (a – 6) – 4 (a – 6) = 0
If (a – 4) (a – 6) = 0
⇒ a = 4, a = 6
If a = 4
⇒ x + \(\frac{1}{x}\) = 4
⇒ x² – 4x + 1 = 0
⇒ x = \(\frac{-(-4) \pm \sqrt{16-4(1)}}{2(1)}\)
= \(\frac{4 \pm \sqrt{12}}{2}\)
= 2 ± √3
If a = 6 x + \(\frac{1}{x}\) = 4
⇒ x² – 6x + 1 = 0
x = \(\frac{-(-6) \pm \sqrt{36-4(1)(1)}}{2(1)}=\frac{6 \pm \sqrt{32}}{2}\)
= 3 ± 2√2
Hence ti roots of the given equation are 2 ± √3, 3 ± 2√2.

Question 15.
Solve the quation 6x⁴ – 35x³ + 62x² – 35x + 6 = 0. [May ’13, ’10]
Solution:
Given equation is 6x⁴ – 35x³ + 62x² – 35x + 6 = 0
This is a standard form of reciprocal equation.
The given equation is an even degree reciprocal equation of class – 1
On dividing both sides of given equation by x², we get
6x² – 35x + 62 – \(\frac{35}{x}+\frac{6}{x^2}\) = 0
\(6\left(x^2+\frac{1}{x^2}\right)-35\left(x+\frac{1}{x}\right)\) + 62 = 0
Let x + \(\frac{1}{x}\) = a
⇒ \(\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\) = a² – 2
⇒ 6(a² – 2) – 35a + 62 = 0
⇒ 6a² – 12 – 35a + 62 = 0
⇒ 6a² – 35a + 50 = 0
⇒ 6a² – 15a – 20a + 50 = 0
⇒ 3a (2a – 5) – 10 (2a – 5) = 0
⇒ (2a – 5)(3a – 10) = 0
⇒ 2a = 5; 3a = 10
a = \(\frac{5}{2}\) a = \(\frac{10}{3}\)
If a = \(\frac{5}{2}\)
⇒ x + \(\frac{1}{x}\) = \(\frac{5}{2}\)
⇒ 2x² + 2 = 5x
⇒ 2x² – 5x + 2 = 0
⇒ 2x² – 4x – x + 2 = 0
⇒ 2x (x – 2) – 1 (x – 2) = 0
⇒ x = \(\frac{1}{2}\) (or) x = 2
If a = \(\frac{10}{3}\)
⇒ x + \(\frac{1}{x}\) = \(\frac{10}{3}\)
⇒ 3x² + 3 = 10x
⇒ 3x² – 10x + 3 = 0
⇒ 3x² – 9x – x + 3 = 0
⇒ 3x (x – 3) – 1 (x – 3) = 0
x = \(\frac{1}{3}\) (or) x = 3
∴ The roots are 2, 3, \(\frac{1}{2}\), \(\frac{1}{3}\).

Question 16.
Solve 2x⁵ + x⁴ – 12x³ – 12x² + x + 2 = 0. [TS – May ‘15: AP-Mar. ’17, ‘16; March ‘08, ‘07]
Solution:
Given equation is 2x⁵ + x⁴ – 12x³ – 12x² + x + 2 = 0
Let f(x) = 2x⁵ + x⁴ – 12x³ – 12x² + x + 2 = 0
If x = – 1
⇒ (- 1) = 2 (- 1) + 1 – 12 (- 1) – 12(1) + (- 1) + 2
= – 2 + 1 + 12 – 12 – 1 + 2 = 0
∴ (x + 1) is a factor of f(x).
On dividing f(x) by (x + 1) we get,

∴ 2x⁴ – x³ – 11x² – x + 2 = 0
On dividing with x² on both sides, we get
⇒ 2x² – x – 11 – \(\frac{1}{x}+\frac{2}{x^2}\) = 0
⇒ \(2\left(x^2+\frac{1}{x^2}\right)-1\left(x+\frac{1}{x}\right)\) – 11 = 0
Let x + \(\frac{1}{x}\) = a
⇒ x² + \(\frac{1}{x^2}\) + 2 = a²
⇒ x² + \(\frac{1}{x^2}\) = a² – 2
⇒ 2(a² – 2) – 1 (a) – 11 = 0
⇒ 2a² – 4 – a – 11 = 0
⇒ 2a² – a – 15 = 0
⇒ 2a² – 6a + 5a – 15 = 0
⇒ 2a (a – 3) + 5 (a – 3) = 0
⇒ (a – 3) (2a + 5) = 0
a = 3; a = – \(\frac{5}{2}\)
If a = 3;
x + \(\frac{1}{x}\) = 3
x² – 3x + 1 = 0
x = \(\frac{-(-3) \pm \sqrt{9-4}}{2(1)}=\frac{3 \pm \sqrt{5}}{2}\)
If a = – \(\frac{5}{2}\)
x + \(\frac{1}{x}\) = – \(\frac{5}{2}\)
⇒ 2x² + 5x + 2 = 0
⇒ 2x² + 4x + x + 2 = 0
⇒ 2x (x + 2) + 1 (x + 2) = 0
⇒ (2x + 1) (x + 2) = 0
∴ x = – \(\frac{1}{2}\), x = – 2
∴ The roots are – 1, – \(\frac{1}{2}\), – 2, \(\frac{3 \pm \sqrt{5}}{2}\).

Question 17.
Solve the equation x⁵ – 5x⁴ + 9x³ – 9x² + 5x – 1 = 0. [AP – May 2016; March ‘13]
Solution:
Given equation is x⁵ – 5x⁴ + 9x³ – 9x² + 5x – 1 = 0
Let f(x) = x⁵ – 5x⁴ + 9x³ – 9x² + 5x – 1
If x = 1
f(1)= 1⁵ – 5(1)⁴ + 9(1)³ – 9(1)² + 5(1) – 1
= 1 – 5 + 9 – 9 + 5 – 1 = 0
∴ (x – 1) is a factor of 1(x).
On dividing x⁵ – 5x³ + 9x³ – 9x² + 5x – 1 by x – 1

∴ x⁴ – 4x³ + 5x² – 4x + 1 = 0
This is a standard form of reciprocal equation.
Now, on dividing with x² on both sides, we get
x² – 4x + 5 – \(\frac{4}{x}+\frac{1}{x^2}\) = 0
\(\left(x^2+\frac{1}{x^2}\right)\) – 4 \(\left(x+\frac{1}{x}\right)\) + 5 = 0
Let x + \(\frac{1}{x}\) = a
⇒ [x + \(\frac{1}{x}\)]² = a²
⇒ x² + \(\frac{1}{x^2}\) + 2 = a²
⇒ x² + \(\frac{1}{x^2}\) = a² – 2
⇒ a² – 2 – 4a + 5 = 0
⇒ a² – 4a + 3 = 0
⇒ a² – 3a – a + 3 = 0
⇒ a (a – 3) – 1 (a – 3) = 0
⇒ (a – 1) (a – 3) = 0
⇒ a = 1 (or) a = 3
If a = 1
⇒ x + \(\frac{1}{x}\) = 1
⇒ x² – x + 1 = 0
⇒ x = \(\frac{-(-1) \pm \sqrt{1-4}}{2(1)}=\frac{1 \pm \mathrm{i} \sqrt{3}}{2}\)
If a = 3
⇒ x + \(\frac{1}{x}\) = 3
⇒ x² – 3x + 1 = 0
⇒ x = \(\frac{-(-3) \pm \sqrt{9-4}}{2(1)}=\frac{3 \pm \sqrt{5}}{2}\)
∴ The roots are 1, \(\frac{1 \pm \mathrm{i} \sqrt{3}}{2}\), \(\frac{3 \pm \sqrt{5}}{2}\).

Question 18.
Solve the equation 6x⁶ – 25x⁵ + 31x⁴ – 31x² + 25x – 6 = 0. [AP – Mar. ‘18. May 2015]
Solution:
Given equation is 6x⁶ – 25x⁵ + 31x⁴ – 31x² + 25x – 6 = 0
The given equation is an even degree reciprocal equation of class – 2.
Let f(x) = 6x⁶ – 25x⁵ + 31x⁴ – 31x² + 25x – 6 = 0
If x = 1 then f(1) = 6 – 25 + 31 – 31 + 25 – 6 = 0
∴ (x – 1) is a factor of f(x).
If x = – 1 then
f(- 1) = 6 + 25 + 31 – 31 – 25 – 6 = 0
∴ (x + 1) is a factor of f(x).
On dividing 1(x) by x + 1 and x – 1 we get,

∴ Quotient is 6x⁴ – 25x³ + 37x² – 25x + 6
6x⁴ – 25x³ + 37x² – 25x + 6 = 0
This is standard form of reciprocal equation.
This is an even degree reciprocal equation of class – 1.
On dividing both sides with x².
6x² – 25x + 37 – \(\frac{25}{x}+\frac{6}{x^2}\) = 0
\(6\left(x^2+\frac{1}{x^2}\right)-25\left(x+\frac{1}{x}\right)\) + 37 = 0
Let x + \(\frac{1}{x}\) = a
\(x^2+\frac{1}{x^2}\) = a² – 2
⇒ 6(a² – 2) – 25a + 37 = 0
⇒ 6a² – 25a + 25 = 0
⇒ 6a² – 15a – 10a + 25 = 0
⇒ 3a(2a – 5) – 5 (2a – 5) = 0
⇒ (3a -5) (2a – 5) = 0
⇒ a = \(\frac{5}{2}\), a = \(\frac{5}{3}\)
If a = \(\frac{5{2}\)
⇒ x + \(\frac{1}{x}\) = a
⇒ x + \(\frac{1}{x}\) = \(\frac{5}{2}\)
⇒ 2x² – 5x + 2 = 0
⇒ 2x² – 4x – x + 2 = 0
⇒ 2x (x – 2) – 1 (x – 2) = 0
⇒ x = \(\frac{1}{2}\); x = 2
If a = \(\frac{5}{3}\)
⇒ x + \(\frac{1}{x}\) = \(\frac{5}{3}\)
⇒ 3x² – 5x + 3 = 0
x = \(\frac{-(-5) \pm \sqrt{25-4(3)(3)}}{2(3)}\)
= \(\frac{5 \pm \sqrt{25-36}}{6}=\frac{5 \pm \mathrm{i} \sqrt{11}}{6}\)
∴ The roots are ± 1, 2, \(\frac{1}{2}\), \(\frac{5 \pm \mathrm{i} \sqrt{11}}{6}\).

Question 19.
Find the polynomial equation of degree 5 whose roots are the translates of the roots of x₅ + 4x₃ – x₂ + 11 = 0 by – 3. [March ’06]
Solution:
Let f(x) = x⁵ + 4x³ – x² + 11 = 0
∴ The required equation is f(x + 3) = 0
∴ f(x + 3) = A₀x⁵ + A₁x⁴ + A₂x³ + A₃x² + A₄x + A₅
The coefficients A₀, A₁, A₂, A₃, A₄, A₅ can be obtained as follows

∴ Required equation is f(x + 3) = 0
x₅ + 15x₄ + 94x³ + 305x₂ + 507x + 353 = 0

Question 20.
Solve the equation x⁴ – 9x³ + 27x² – 29x + 6 = 0 given that one root is 2 – √3. [May ’03]
Solution:
Given equation is x⁴ – 9x³ + 27x² – 29x + 6 = 0
2 – √3 is a root of it then 2 + √3 is also root.
The quadratic factor of these two roots is (x – (2 + √3)) (x – (2 – √3))
= (x – (2 – √3)) (x – (2 + √3))
= ((x – 2) – √3) ((x – 2) + √3)
= (x – 2)² – (√3)²
= x² – 4x + 4 – 3
= x² – 4x + 1
On dividing x⁴ – 9x³ + 27x² – 29x + 6 with x² – 4x + 1

∴ Quotient is x² – 5x + 6 = 0
∴ x² – 5x + 6 = 0
⇒ x² – 3x – 2x + 6 = 0
⇒ x(x – 3) – 2 (x – 3) = 0
⇒ (x – 2) (x – 3) = 0
⇒ x = 2 (or) x = 3
∴ The roots of given equation are 2, 3, 2 ± √3.

Question 21.
Show that x⁵ – 5x³ + 5x² – 1 = 0, has three equal roots and find that root. [TS – Mar. 2017]
Solution:
Given equation is x⁵ – 5x³ + 5x² – 1 = 0
Let f(x) = x⁵ – 5x³ + 5x² – 1
f'(x) = 5x⁴ – 15x² + 10x
f”(x) = 20x³ – 30x + 10
f(1) = 20 – 30 + 10 = 0
Similarly f'(1) = 0 and f(1) = 0
∴ (x – 1) is a factor of f”(x), f'(x) and f(x).
Thus f(x) = 0 has three equal roots and it is 1.

Question 22.
Find the condition that x³ – px² + qx – r = 0 may have the roots in G.P.
Solution:
q³ = p³r.

Question 23.
Solve 8x³ – 36x² – 18x + 81 = 0, given that the roots are in AP. [March ‘04]
Solution:
\(\frac{-3}{2}, \frac{3}{2}, \frac{9}{2}\).

Question 24.
Solve 3x³ – 26x² + 52x – 24 = 0, given that the roots are in A.P.
Solution:
\(\frac{2}{3}\), 2, 6

Question 25.
Solve the equation 6x³ – 11x² + 6x – 1 = 0, the roots being in H.P.
Solution:
1, \(\frac{1}{2}\), \(\frac{1}{3}\)

Question 26.
Solve x³ – 7x² + 36 = 0, gi0ven one root being twice the other.
Solution:
3, 6, – 2.

Question 27.
Solve x⁴ – 5x³ + 5x² + 5x – 6 = 0, given that the product of two of its roots ¡s 3.
Solution:
2, – 1, 3, 1

Question 28.
Solve 9x³ – 15x² + 7x – 1 = 0, given that two of its roots are equal.
Solution:
\(\frac{1}{3}\), \(\frac{1}{3}\), 1.

Question 29.
Solve x³ – 3x² – 10x + 48 = 0.
Solution:
3, – 4, 4.

Question 30.
Solve the equation 4x³ – 13x² – 13x + 4 = 0.
Solution:
– 1, \(\frac{1}{4}\), – 4

Question 31.
Find the polynomial equation whose roots are the translates of those of the equation. [TS – Mar. ’16]
x⁵ – 4x⁴ + 3x² – 4x + 6 = Oby – 3.
Solution:
x⁵ + 11x⁴ + 42x³ – 57x² – 13x – 60 = 0

Question 32.
Solve the equation x⁴ + 2x³ – 16x² – 22x + 7 = 0 given that 2 – √3 is one of its roots.
Solution:
2 ± √3, – 3 ± √2.

Some More Maths 2A Theory of Equations Important Questioons

Question 1.
Form the monic polynomial equation of degree 3 whose roots are 2, 3 and 6. [March ‘02]
Solution:
Let α = 2, β = 3, γ = 6
The monk equation having roots α, β, γ is
(x – α) (x – β) (x – γ) = 0
⇒ (x – 2) (x – 3) (x – 6) = 0
⇒ (x – 2) (x² – 9x + 18) =0
⇒ x³ – 9x² + 18x – 2x² + 18x – 36 = 0
⇒ x³ – 11x² + 36x – 36 = 0.

Question 2.
Form the monic polynomial equation of degree 4 whose roots are 4 + √3, 4 – √3, 2 + i and 2 – i.
Solution:
Let α = 4 + √3, β = 4 – √3, γ = 2 + i, δ = 2 – i
The equation whose roots are α, β, γ, δ is
(x – α) (x – β) (x – γ) (x – δ) = 0
(x – (4 + √3)) (x – (4 – √3))
(x – (2 + i)) (x – (2 – i)) = 0
((x – 4) – √3) ((x – 4) + √3)
((x – 2) – i) ((x – 2) + i) = 0
((x – 4)² – 3) ((x – 2)² – 2) = 0
(x² + 16 – 8x – 3) (x² + 4 – 4x + 1) = 0
(x² – 8x + 13) (x² – 4x + 5) = 0
x⁴ – 12x³ + 50x² – 92x + 65 = 0

Question 3.
Find s₁, s₂, s₃ and s₄ for the equation 8×4 – 2×3 – 27×2 – 6x + 9=0.
Solution:
Given equation is 8x⁴ – 2x³ – 27x² – 6x + 9 = 0
Comparing this equation with ax⁴ + bx³ + cx² + dx + e = 0
a = 8; b = – 2; c = – 27; d = – 6; e = 0
Now,
s₁ = \(\frac{-b}{a}=\frac{-(-2)}{8}=\frac{1}{4}\)
s₂ = \(\frac{c}{a}=\frac{-27}{8}\)
s₃ = \(\frac{-\mathrm{d}}{\mathrm{a}}=\frac{-(-6)}{8}=\frac{3}{4}\)
s₄ = \(\frac{\mathrm{e}}{\mathrm{a}}=\frac{9}{8}\)

Question 4.
Find the algebraic equation whose roots are 3 times the roots of x³ + 2x² – 4x + 1 = 0.
Solution:
Let f(x) = x³ + 2x² – 4x + 1 = 0
∴ Required equation is f(\(\frac{x}{3}\)) = 0
\(\) = 0
⇒x3 + 6×2-36x + 27=0.

Question 5.
Find an algebraic equation of degree 4 whose roots are 3 times the roots of the equation 6x⁴ – 7x³ + 8x² – 7x + 2 = 0.
Solution:
Let f(x) = 6x⁴ – 7x³ + 8x² – 7x + 2 = 0
The required equation is f(\(\frac{x}{3}\)) = 0
\(\frac{6 x^4}{81}-\frac{7 x^3}{27}+\frac{8 x^2}{9}-\frac{7 x}{3}+2\) = 0
6x⁴ – 21x³ + 72x² – 189x + 162 = 0

Question 6.
Find the transfonned equation whose roots are the negatives of the roots of x⁴ + 5x³ + 11x + 3 = 0.
Solution:
Let f(x) = x⁴ + 5x³ + 11x + 3 = 0
Required equation ¡s f(- x) = 0
⇒ (- x)⁴ + 5(- x)³ + 11(- x) + 3 = 0
⇒ x⁴ – 5x³ – 11x + 3 = 0

Question 7.
Find the polynomial equation of degree 4 whose roots are the negatives of the roots of x⁴ – 6x³ + 7x² – 2x + 1 = 0.
Solution:
Let f(x) = x⁴ – 6x³ + 7x² – 2x + 1 = 0
The required equation is f(- x) = 0
⇒ (- x)⁴ – 6(- x)³ + 7(- x)² – 2(- x) + 1 = 0
⇒ x⁴ + 6x³ + 7x² + 2x + 1 = 0.

Question 8.
Find the polynomial equation whose roots are the reciprocals of the roots of x⁵ + 11x⁴ + x³ + 4x² – 13x + 6 = 0
Solution:
Let f(x) = x⁵ + 11x⁴ + x³ + 4x² – 13x + 6 = 0
The required equation is f(\(\frac{1}{x}\)) = o
⇒ \(\frac{1}{x^5}+\frac{11}{x^4}+\frac{1}{x^3}+\frac{4}{x^2}-\frac{13}{x}+6\) + 6 = 0
⇒ 1 + 11x + x² + 4x³ – 13x⁴ + 6x⁵ = 0
⇒ 6x⁵ – 13x⁴ + 4x³ + x² + 11x + 1 = 0

Question 9.
Find the polynomial equation whose roots are the reciprocals of the roots of the equation x⁴ + 3x³ – 6x² + 2x – 4 = 0.
Solution:
Let f(x) = x⁴ + 3x³ – 6x² + 2x – 4 = 0
The required equation is f(\(\frac{1}{x}\))) = 0
⇒ \(\frac{1}{x^4}+\frac{3}{x^3}-\frac{6}{x^2}+\frac{2}{x}-4\) = 0
⇒ 4x⁴ – 2x³ + 6x² – 3x – 1 = 0

Question 10.
Find the polynomial equation whose roots are the squares of the roots of x⁴ + x³ + 2x² + x + 1 = 0.
Solution:
Let f(x) = x⁴ + x³ + 2x² + x + 1 = 0
The required equation is f(√x) = 0
(√x)⁴ + (√x)³ + 2(√x)² + √x + 1 = 0
x² + x√x + 2x + √x + 1 = 0
x² + 2x + 1 = – x(√x) – √x
= – √x (x + 1)
⇒ squaring on both sides
(x² + 2x + 1)² = (- √x (x + 1))²
⇒ x⁴ + 4x² + 1 + 4x³ + 4x + 2x² = x(x² + 1 + 2x) = x³ + x + 2x²
x⁴ + 3x³ + 4x² + 3x + 1 = 0.

Question 11.
Find the polynomial equation whose roots are the squares of the roots of x³ – x² + 8x – 6 = 0.
Solution:
Let f(x) = x³ – x² + 8x – 6 = 0
∴ The required equation is f(√x) = o
(√x)³ – (√x)² + 8√x + 6 = 0
⇒ x√x – x + 8√x – 6 = 0
⇒ x + 6 = √x (x + 8)
⇒ squaring on both sides
⇒x² + 36 + 12x = x(x² + 64 + 16x)
⇒ x³ + 64x + 16x²
⇒ x³ + 15x² + 52x – 36 = 0.

Question 12.
Find the condition that x³ – px² + qx – r = 0 may have the roots in G.P.
Solution:
Given equation is x³ – px² + qx – r = 0
Since the roots are in G.P. then they must be the of the form \(\frac{a}{r}\), a, ar.
Now s₃ = \(\frac{a}{r}\) . a . ar
= \(\frac{-(-r)}{1}\)
a³ = r
a = r^1/3
Since ‘a’ is a root of x³ – px² + qx – r = 0
(r^1/3)³ – p(r^1/3)² + q(r^1/3) . r = 0.
⇒ r – pr^2/3 + qr^1/3 – r = 0
⇒ pr^2/3 = qr^1/3
Cubing on both sides.
⇒ p³r² = q³r
⇒ p³r = q³

Question 13.
Solve 8x³ – 36x² – 18x + 81 = 0, given that the roots are in AP.
Solution;
Given equation is 8x³ – 36x² – 18x + 81 = 0
Since, the roots are in A.P, they must be of the form a – d, a, a + d.
Now, a – d + a + a + d = \(\frac{-(-36)}{8}\)
⇒ 3a = \(\frac{9}{2}\)
⇒ a = \(\frac{3}{2}\)
(a – d) a (a + d) = \(\frac{-81}{8}\)
(a² – d²) a = \(\frac{-81}{8}\)
\(\frac{3}{2}\left(\frac{9}{4}-\mathrm{d}^2\right)=\frac{-81}{8}\)
\(\frac{9}{4}-d^2=\frac{-27}{4}\)
d² = 36
⇒ d = ± 3.

Case – I:
a = \(\frac{3}{2}\), d = 3
The roots of the given equation are a – d, a, a + d
\(\frac{-3}{2}, \frac{3}{2}, \frac{9}{2}\)

Case – II: a = \(\frac{3}{2}\), d = – 3
The roots of the given equation are a – d, a, a + d
\(\frac{9}{2}, \frac{3}{2}, \frac{-3}{2}\)

Question 14.
Solve 3x³ – 26x² + 52x – 24 = 0 the given, the roots are in Geometric Progression.
Solution:
Given equation is 3x³ – 26x² + 52x – 24 = 0
Since the roots are in G.P they must be of the form \(\frac{a}{r}\), a, ar.
Now, \(\frac{a}{r}\) + a + ar = \(\frac{-(-26)}{3}\)
a(\(\frac{1}{r}\) + 1 + r) = \(\frac{26}{3}\) …………..(1)
\(\frac{a}{r}\) . a . ar = \(\frac{-(-24)}{3}\)
a³ = 8
a = 2
From (1)
⇒ \(2\left(\frac{1+r+r^2}{r}\right)=\frac{26}{3}\)
⇒ 3(r² + r + 1) = 13r
⇒ 3r² – 10r + 3 = 0
⇒ 3r² – 9r – r + 3 = 0
⇒ 3r (r – 3) – 1(r – 3) = 0
⇒ (r – 3) (3r – 1) = 0
r = 3 (or) r = \(\frac{1}{3}\)

Case – 1:
a = 2, r = 3
The roots of the given equation are \(\frac{a}{r}\), a, ar = \(\frac{2}{3}\), 2, 6

Case – II:
a = 2, r = \(\frac{1}{3}\)
The roots of the given equation are \(\frac{a}{r}\), a, ar = 6, 2, \(\frac{2}{3}\).

Question 15.
Solve the equation 6x³ – 11x² + 6x – 1 = 0, the roots being in 1-LP.
Solution:
Given equation is 6x³ – 11x² + 6x – 1 = 0
The roots of 6x³ – 11x² + 6x – 1 = 0 are in H.P
The roots of \(6 \cdot \frac{1}{x^3}-11 \cdot \frac{1}{x^2}+\frac{6}{x}-1\) = 0 are in A.P.
6 – 11x + 6x² – x³ = 0
x³ – 6x² + 11x – 6 = 0
Let the roots be a – d, a, a + d
a – d + a + a + d = \(\frac{-(-6)}{1}\)
3a = 6
a = 2
(a – d) a (a + d) = \(\frac{-(-6)}{1}\) = 6
⇒ 2(a² – d²) = 6
⇒ 4 – d² = 3
⇒ d² = 1
⇒ d = 1
The roots of x³ – 6x² + 11x – 6 = 0 are a – d, a, a + d = 1, 2, 3.
∴ The roots of 6x³ – 11x² + 6x – 1 = 0 are 1, \(\frac{1}{2}\), \(\frac{1}{3}\).

Question 16.
Solve x³ – 7x² + 36 = 0, given one root being twice the other.
Solution;
Given equation is x’ – 7×2 + 36 = 0
Let α, β, γ be the roots of the given equation.
Now, α + β + γ = \(\frac{-(-7)}{1}\)
⇒ α + β + γ = 7 …………..(1)
αβ + βγ + γα = \(\frac{0}{1}\) = 0
⇒ αβ + βγ + γα = 0 ………….(2)
αβγ = \(\frac{-36}{1}\) = – 36
⇒ αβγ = – 36 ……………(3)
Given that one root being twke the other then β = 2α
(1) ⇒ α + 2α + γ = 7
⇒ 3α + γ = 7
⇒ α = 7 – 3α
(2) ⇒ α . 2α + 2αγ + αγ = 0
⇒ 2α + 3αγ = o
⇒ 2α² + 3α (7 – 3α) = 0
⇒ 2α² + 21α – 9α² = 0
⇒ 7α² – 21α = 0
⇒ α² – 3α =0
⇒ α (α – 3) = 0
⇒ α = 0 (or) α = 3
α = 0 does not satisfy the given equation.
∴ α = 3
If α =3
⇒ β = 6
γ = 7 – 3α
= 7 – 3 . 3
= 7 – 9
⇒ γ = – 2
∴ The roots of the given equation are 3, 6, – 2.

Question 17.
Solve x⁴ – 5x³ + 5x² + 5x – 6 = 0, given that the product of two of its roots is 3.
Solution:
Let α, β, γ, δ are the roots of given equation.
⇒ αβγδ = – 6
Given that the product of two roots is 3
αβ = 3
⇒ 3γδ = – 6
⇒ γδ = – 2
Let α + β = p, γ + δ = q
The quadratic equation whose roots are α, β is
x² – x(α + β) + αβ = 0
⇒ x² – px + 3 = 0 …………….(1)
The quadratic equation ol roots γ, δ is
x² – x(γ + δ) + γδ = 0
⇒ x² – qx – 2 = 0 ………………(2)
∴ x⁴ – 5x³ + 5x² + 5x – 6 = (x² – px + 3) (x² – qx – 2)
= x⁴ – qx³ – 2x² – px³ + pqx² + 2px + 3x² – 3qx – 6
= x⁴ + x ³ (- p – q) + x² (1 + pq) + x (2p – 3q) – 6
x³coeff. ⇒ – p – q = – 5
⇒ p + q = 5 ……………..(3)
xcoeff. ⇒ 2p – 3q = 5 …………..(4)
Solving (3) and (4)

p = 4, q = 1
Sub. p = 4 in (1)
⇒ x² -4x + 3 = 0
⇒ x² – 3x – x + 3 = 0
⇒ x (x – 3) – 1 (x – 3) = 0
⇒ (x – 1) (x – 3) = 0
⇒ x – 1 = 0 (or) x – 3 = 0
⇒ x = 1 (or) x = 3
Sub.q = 1 in (2)
⇒ x² – x – 2 = 0
⇒ x² – 2x + x – 2 = 0
⇒ x (x – 2) + 1 (x – 2) = 0
⇒ (x + 1) (x – 2) = 0
⇒ x = 2 (or) x = – 1
The roots of given equation are – 1, 1, 2, 3.

Question 18.
Solve 9x³ – 15x² + 7x – 1 = 0, given that two of its roots are equal.
Solution:
Given equation is 9x³ – 15x² + 7x – 1 = 0
Let α, β, γ are the roots of 9x – 15x² + 7x – 1 = 0
Now, α + β + γ = \(\frac{-(-15)}{9}=\frac{5}{3}\) ⇒ 1
⇒ αβ + βγ + γα = \(\frac{7}{9}\) …………….(2)
αβγ = \(\frac{-(-1)}{9}=\frac{1}{9}\) …………(3)
Given α = β; (1)
⇒ 2α + γ = \(\frac{5}{3}\) ……………(4)
(2) ⇒ α² + αγ + γα = \(\frac{7}{9}\)
α² + 2αγ = \(\frac{7}{9}\)
γ = \(\frac{7}{18 \alpha}-\frac{\alpha}{2}\) ………….(5)
From (4) 2α + \(\frac{7}{18 \alpha}-\frac{\alpha}{2}=\frac{5}{3}\)
⇒ \(\frac{36 \alpha^2+7-9 \alpha^2}{18 \alpha}=\frac{5}{3}\)
⇒ 27α² – 30α + 7 = 0
⇒ 27α² – 21α – 9α + 7 = 0
⇒ 3α (9α – 7) – 1 (9α – 7) = 0
⇒ (3α – 1) (9α – 7) = 0
∴ α = \(\frac{1}{3}\), β = \(\frac{7}{9}\)
α = does not satisfy the given equation.
∴ α = \(\frac{1}{3}\), β = \(\frac{1}{3}\)
(4) ⇒ \(\frac{2}{3}\) + γ = \(\frac{5}{3}\)
γ = \(\frac{5}{3}-\frac{2}{3}=\frac{3}{3}\) = 1
γ = 1
∴ The roots of the given equation are \(\frac{1}{3}\), \(\frac{1}{3}\), 1.

Question 19.
Solve x³ – 3x² – 16x + 48 = 0.
Solution:
Let f(x) = x³ – 3x² – 16x + 48 = 0
If x = 3
⇒ f(3) = 27 – 27 – 48 + 48 = 0
Hence 3 is a root of f(x) = 0.
Now, we divide f(x) by x – 3 using synthetic division

∴ The quotient is x² – 16 and the remainder is zero.
∴ f(x) = x³ – 3x² – 16x + 48
= (x – 3) (x² – 16)
= (x – 3) (x + 4) (x – 4)
∴ The roots are 3, – 4, 4.

Question 20.
Solve the equation 4x³ – 13x² – 13x + 4 = 0.
Solution:
Given equation is 4x³ – 13x² – 13x 4 = 0.
f(x) = 4x³ – 13x² – 13x + 4
If x = – 1
⇒ f(- 1) = 4(- 1)³ – 13(- 1)² – 13(- 1) + 4
= – 4 – 13 + 13 + 4 = 0
∴ (x + 1) is a factor of f(x).
Now, on dividing 4x³ – 13x² – 13x + 4 = 0 by x + 1.

Quotient is 4x² – 17x + 4
4x² – 17x + 4 = 0
4x² – 16x – x + 4 = 0
=4x (x – 4) – 1 (x – 4) = 0
(4x – 1) (x – 4) = 0
x = \(\frac{1}{4}\), x = 4
∴ 4x³ – 13x² – 13x + 4 = (x + 1) (x – 4) (4x – 1)
∴ The roots of the given equation are – 1, 4, 1/4.

Question 21.
Find the polynomial equation whose roots are the translates of those of the equatior x⁵ – 4x⁴ + 3x² – 4x + 6 = 0 by – 3.
Solution:
Let f(x) = x⁵ – 4x⁴ + 3x² – 4x + 6
∴ The required equation is f(x + 3) =0
∴ f(x+3) = A₀x⁵ + A₁x⁴ + A₂x³ + A₃x² + A₄x + A₅
The coefficients A₀, A₁, A₂, A₃, A₄, A₅ can be obtained as follows.

∴ Required equation ¡s f(x + 3) = 0
x⁵ + 11x⁴ + 42x³ + 57x² – 13x – 60 = 0.

Question 22.
Solve the equation x⁴ + 2x³ – 16x² – 22x + 7 = 0 given that 2 – √3 is one of its roots.
Solution:
Given equation is x⁴ + 2x³ – 16x² – 22x + 7 = 0 ……………..(1)
Let f(x) = x⁴ + 2x³ – 16x² – 22x + 7
Given 2 – √3 is a root of (1).
⇒ 2 + √3 is also a root of (1).
(∵ coefficients of (1) are rational)
∴ (x – (2 – √3)) (x – (2 + √3)) is a factor of f(x).
(x² – 4x + 1)is a factor of f(x).
We divide f(x) by x² – 4x 1.
By synthetic division,

∴ f(x)=(x² – 4x +1 ) (x² + 6x + 7)
∴ Equation (1)
z(x² – 4x + 1) (x² + 6x + 7) = 0
=x² – 4x + 1 = 0 or x² + 6x + 7 = 0
x = 2 ± √3 or x = – 3 ± √2
∴ The roots of the given equation are 2 ± √3, – 3 ± √2.

Question 23.
If α, β, γ are the roots of x³ + px² + qx + r = 0, then form the monic cubic equation whose roots are α(β + γ), β(γ + α), γ(α + β).
Solution:
Given equation is x³ + px² + qx + r = 0
Since α, β, γ are the roots of the equation.
x³ + px² + qx + r = 0 then
s₁ = α + β + γ = \(\frac{-p}{1}\) = – p;
s₂ = αβ + βγ + γα = \(\frac{q}{1}\) = q;
s₃ = αβγ = \(\frac{-r}{1}\) = – r
Let s₁ = Σ α(β + γ)
= Σ (αβ + αγ) = Σ αβ + Σ αγ = Σ αβ + Σ αβ = 2Σ αβ = 2q
s₂ = Σ α(β + γ) . β(γ + α)
= Σ αβ (- p – α) (- p – β)
= Σ αβ (p + α) (p + β)
= Σ αp (p² + p(α + β) + αβ)
= Σ αp [p² + p(- p – γ) + ap]
= Σ αβ [p² – p² – pγ + αβ]
= – pΣ αβγ + Σ αβ
= – p(3αβγ) + (αβ + βγ + γα)² – 2αβγ(α + β + γ)
= – 3p (- r) + q² – 2 (- r) (- p)
= 3pr + q² – 2pr
= q² + pr
s₃ = α (β + γ) β (γ + α) γ (α + β)
= αβγ(- p – α) (- p – β) (- p – γ)
= – αβγ [(p + α) (p + β) (p + γ)]
= – αβγ [(p² + pα + pβ + αβ) (p + γ)]
= – αβγ [p³ + p²α + p²β + pαβ + p²γ + pαγ + pβγ + αβγ]
= – αβγ [p³ + p² (α + β + γ) + p(αβ + βγ + γα) + αβγ]
= – (- γ)[p³ + p² (- p) + p(q) + (- r)]
= γ [p³ – p³ + pq – r]
= pqr – r²
∴ The monic cubic equation whose roots are α (β + γ), β(γ + α), γ(α + β) is
x³ – s₁x² + s₂x – s₃ = 0
x³ – 2qx² + (q² + pr) x – r (pq – r) = 0.

Question 24.
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:
Let f(x) = x³ – 6x² + 11x – 6
If x = 1 then f(1) = 1 – 6 + 11 – 6 = 12 – 12 = 0
(x – 1) is the factor of f(x)

x² – 5x + 6 = 0
⇒ x² – 3x – 2x + 6 = 0
⇒ x (x – 3) – 2 (x – 3) = 0
⇒ (x – 2) (x – 3) = 0
⇒ x – 2 = 0; x – 3 = 0
Therefore, x³ – 6x² + 11x – 6 = 0
⇒ (x – 1) (x – 2) (x – 3) = 0
⇒ x = 1, 2, 3
Since α, β, γ are the roots of x³ – 11x² + 11x – 6 = 0
Then, α = 1; β = 2; γ = 3.
Now, α² + β² = 1 + 4 = 5;
β² + γ² = 4 + 9 = 13;
γ² + α² = 9 + 1 = 10
The equation having roots 5, 13, 10 is (x – 5) (x – 13) (x – 10) = 0
= (x² – 18x + 65) (x – 10) = 0
x³ – 18x² + 65x – 10x² + 180x – 650 = 0
x³ – 28x² + 245x – 650 = 0

Question 25.
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:
Let f(x) = x³ – 7x + 6
If x = 1 then f(1) = 1 – 7(1) + 6 – 7 – 7 = 0
∴ (x – 1) is a factor of f(x)
x² + x – 6 = 0
⇒ x² + 3x – 2x – 6 = 0

x (x + 3) – 2 (x + 3) = 0
(x + 3) (x – 2) = 0
∴ x³ – 7x + 6 = 0
⇒ (x + 3) (x – 1) (x – 2) = 0
⇒ x = – 3, 1, 2
Since α, β, γ are the roots of x³ – 7x + 6 = 0
then, α = – 3; β = 1; γ = 2.
Now, (α – β)² = (- 3 – 1)² = 16;
(β – γ)² = (1 – 2)² = 1;
(γ – α)² = (4 + 9)² = 25
The equation having roots 16, 1, 25 is (x – 16) (x – 1) (x – 25) = 0
(x² – 1 7x + 16) (x² – 25) = 0
x³ – 17x² + 16x – 25x² + 465x – 400 = 0
x³ – 42x² + 441x – 400 = 0.

Question 26.
Solve x³ – 9x² + 14x + 24 = 0 given that two of the roots are in the ratIo 3: 2.
Solution:
Let the roots be 3α, 2α, β
3α + 2α + β = \(\frac{-(-9)}{1}\) = 9
⇒ 5α + β = 9
⇒ β = 9 – 5α ……………..(1)
3α . 2α + 2αβ + 3αβ = \(\frac{14}{1}\)
⇒ 6α² + 5αβ = 14
6α² + 5α – 14 = 0
⇒ 6α² + 5α (9 – 5α)-14 =0
6α² + 45α – 25α – 14 = 0
⇒ 19α² + 45α – 14 = 0
⇒ 19α² – 45α + 14 = 0
⇒ 19α² – 38α – 7α + 14 = 0
⇒ 19α(α – 2) – 7(α – 2) = 0
(α – 2) (19α – 7) = 0
α = 2 (or)
α = \(\frac{7}{19}\)
⇒ α = \(\frac{14}{19}\)
⇒ 2α = does not satisfy the given equation.
∴ α = 2
3α = 6
2α = 4
(1) ⇒ β = 9 – 5(2)
(2) ⇒ 9 – 10 = – 1
β = – 1
∴ The roots of the given equation are 6, 4,- 1.

Question 27.
Solve the equation x⁴ – 6x³ + 13x² – 24x + 36 = 0 given that they have multiple roots.
Solution:
Let f(x) = x⁴ – 6x³ + 13x² – 24x + 36 = 0
if x = 3 then 81 – 6(27) + 13(9) – 24 (3) + 36
= 81 – 162 + 117 – 72 + 36
= – 234 + 234 = 0
∴ x = 3 is a root of f(x) = 0
Let g(x) = x³ – 3x² + 4x – 12 = 0
If x = 3 then g(3) = 3³ – 3(3²) + 4(3) – 12
= 27 – 27 + 12 – 12 = 0

Let, h(x) = x² + 4 = 0
⇒ x² = – 4
x = ± √4 = ± 2
The roots of the given equation are 3, 3, ± 2i.
Multiple root is 3.

Question 28.
Solve the equation 8x³ – 20x² + 6x + 9 = 0 given that the equation has multiple roots.
Solution:
Given equation is 8x³ – 20x² + 6x + 9 = 0 ………………(1)
Let f(x) = 8x³ – 20x² + 6x + 9
f’(x) = 24x² – 40x + 6
= 2(12x² – 20x + 3)
= 2 (2x – 3) (6x – 1)
\(f\left(\frac{3}{2}\right)=8\left(\frac{27}{8}\right)-20\left(\frac{9}{4}\right)+6\left(\frac{3}{2}\right)+9\)
= 27 – 45 + 9 + 9 = 0
∴ f(\(\left(\frac{3}{2}\right)\)) = 0
∴ f(x) and f’(x) has a common factor ‘2x – 3’.
∴ \(\frac{3}{2}\) is a multiple root of f(x) = 0.
By synthetic division,

∴ 8x³ – 20x² + 6x + 9 = 0
=(x – \(\frac{3}{2}\))² (8x + 4) = 0
x = \(\frac{3}{2}\) or x = \(\frac{-1}{2}\)
∴ The roots of given equation are \(-\frac{1}{2}, \frac{3}{2}, \frac{3}{2}\).

Question 29.
Solve 6x⁴ – 13x³ – 35x² – x + 3 = 0, given that one of its roots is 2 + √3.
Solution:
Given equation is 6x⁴ – 13x³ – 35x² – x + 3 = 0
Since 2 + √3 is a root of it, 2 – √3 is also a root.
The quadratic factor of these two roots is (x – (2 + √3)) (x – (2 – √3))
= ((x – 2) – √3) ((x – 2) + √3)
= (x – 2)² – 3
= x² + 4 – 4x – 3
= x² – 4x + 1
On dividing, 6x⁴ – 13x³ – 35x² – x + 3 by x² – 4x + 1.

∴ Quotent is 6x² + 11x + 3
6x² + 11x + 3 = 0
6x² + 9x + 2x + 3 = 0
3x (2x + 3) + 1 (2x + 3) = 0
= (3x + 1) (2x + 3) = 0
x = \(\frac{-1}{3}\), x = \(\frac{-1}{2}\)
∴ The roots of the given equation are \(\frac{-1}{2}\), \(\frac{-1}{2}\), 2 ± √3.

Question 30.
Solve the equation x⁴ + 2x³ – 5x² + 6x + 2 = 0 given that 1 + i is one of its roots.
Solution:
Given equation is x⁴ + 2x³ – 5x² + 6x + 2 = 0
Since 1 + i is a root of the given equation
then 1 – i is also a root of it.
The quadratic factor to these two roots is
(x – (1 + i)) ((x – (1 – i)) = 0
((x – 1) – i) ((x – 1) + i) = (x – 1) – i
x² + 1 – 2x + 1 = x⁴ – 2x + 2
On dividing x⁴ + 2x³ – 5x² 6x + 2 by x² – 2x + 2

Quotient is x² + 4x + 1
x² + 4x + 1 = 0
x = \(\frac{-4 \pm \sqrt{1-4}}{2(1)}\) = – 2 ± √3
∴ The roots of the given equation are 1 + i, 1 – i, – 2 ± √3.

Question 31.
Solve x⁴ – 4x² + 8x + 35 = 0, given that 2 + √3 is a root. [AP – Mar. 2015]
Solution:
Given equation is x⁴ – 4x² + 8x + 35 = 0
Since 2 + i√3 is a root of it,
2 – i√3 is also a root.
The quadratic factor to these two roots is (x – (2 + i√3)) (x – (2 – i√3))
= ((x – 2) – i√3) ((x – 2) + i√3)
= (x – 2)² (i√3)²
= x² + 4 – 4x + 3
= x² – 4x + 1
On dividing x⁴ – 4x² + 8x + 35 with x² – 4x + 7 we get,

∴ Quotient is x² + 4x + 5
x² + 4x + 5 = 0
x = \(\frac{-4 \pm \sqrt{16-20}}{2(1)}=\frac{-4 \pm \sqrt{-4}}{2}\) = – 2
∴ The roots of the given equation are 2 ± i√3, – 2 ± i.

Question 32.
Solve the equation x⁴ – 6x³ + 11x² – 10x + 2 = 0, given that 2 + √3 is a root of the equation.
Solution:
Given equation is x⁴ – 6x³ + 11x² – 10x + 2 = 0
Since 2 + √3 is a root of it, 2 – √3 is also a root.
The quadratic factor to these roots is (x – (2 + √3)) (x – (2 – √3))
= ((x – 2) – √3) ((x – 2) + √3)
= (x – 2)² – 3
= x² + 4 – 4x – 3
= x² – 4x + 1

∴ Quotient is x² – 2x + 2
∴ x² – 2x + 2 = 0
x = \(\frac{-(-2) \pm \sqrt{4-4(2)}}{2(1)}\)
= \(\frac{2 \pm 2 \mathrm{i}}{2}\) = 1 ± i.
∴ Roots of given equation are 2 ± √3, 1 ± i.

Question 33.
Find the polynomial equation whose roots are the translates of those of the equation x⁴ – 5x³ + 7x² – 17x + 11 = 0 by – 2. [TS – May 2016]
Solution:
Let f(x) = x⁴ – 5x³ + 7x² – 17x + 11 = 0
∴ The required equation is f(x + 2) = 0
Now, f(x + 2) = A₀x⁴ + A₁x³ + A₂x² + A₃x + A₄
Then, the coefficients A₀, A₁, A₂, A₃, A₄ can
be obtained as follows

∴ Required equation is f(x + 2) = 0
x⁴ + 3x³ + x² – 17x – 19 = 0.

Question 34.
Find the polynomial equation whose roots are the translates of the roots of the equation x⁴ – x³ – 10x² + 4x + 24 = 0 by 2.
Solution:
Let f(x) = x⁴ – x³ – 10x² + 4x + 24 = 0
∴ Required equation is f(x -2) = 0
Let f(x – 2) = A₀x⁴ + A₁x³ + A₂x² + A₃x + A₄.
The coefficients A₀, A₁, A₂, A₃, A₄ can be obtained as follows

∴ Required equation is f(x – 2) = 0
x⁴ – 9x³ + 40x² – 80x + 80 = 0.