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

## TS Inter Second Year Maths 2A Theory of Equations Important Questions Very Short Answer Type

Question 1.

Form the polynomial equation of the lowest degree whose roots are 1, – 1, 3. [May ’06]

Solution:

Let α = 1; β = – 1, γ = 3

The equation having roots α, β, γ is

(x – α) (x – β) (x – γ) = 0

⇒ (x – 1) (x + 1) ( x – 3) = 0

⇒ (x^{2} – 1) (x – 3) = 0

⇒ x^{3} – 3x^{2} – x + 3 = 0.

Question 2.

Form the polynomial equation of the lowest degree whose roots are 2 ± √3, 1 ± 2i. [May ’02, ’01]

Solution:

Let

α = 2 + √3, β = 2 – √3, γ = 1 + 2i, δ = 1 – 2i

The equation having α, β, γ, δ is

(x – α) (x – β) (x – γ) (x – δ) = 0

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

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

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

⇒ (x^{2} – 4x + 1) (x^{2} – 2x + 5) = 0

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

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

Question 3.

If 1, 1, α are the roots of x^{3} – 6x^{2} + 9x – 4 = 0. then find α. [AP – Mar. ’18; TS – May 2016; May ’11]

Solution:

Given equation is x^{3} – 6x^{2} + 9x – 4 = 0

Comparing this equation with ax^{3} + bx^{2} + cx + d = 0

we get, a = 1; b = – 6; c = 9; d = – 4

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

then sum of the roots = s_{1} = \(\frac{-b}{a}\)

⇒ 1 + 1 + α = \(\frac{-(-6)}{1}\) = 6

⇒ 2 + α = 6

⇒ α = 4.

Question 4.

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

[AP-Mar. 18; Mar. ’14, ’13, ’10, ’06, May’ 12, ’10]

Solution:

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

Comparing this with ax^{3} + bx^{2} + cx + d = 0 we get,

a = 2; b = 1; c = – 7; d = – 6

Since – 1, 2, α are the roots of 2x^{3} + x^{2} – 7x – 6 = 0

The sum of the roots = s_{1} = \(\frac{-b}{a}\)

– 1 + 2 + α = – \(\frac{1}{2}\)

1 + α = – \(\frac{1}{2}\)

⇒ α = – 1 – \(\frac{1}{2}\)

α = – \(\frac{3}{2}\)

Question 5.

If 1, – 2 and 3 are the roots of x^{3} – 2x^{2} + ax + 6 = 0, then find ‘a’. [TS – May 2015; March ‘04]

Solution:

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

Since 1, – 2, 3 are the roots of x^{3} – 2x^{2} + ax + 6

Now 1 is a root of given equation then

1^{3} – 2(1)^{2} + a(1) + 6 = 0

⇒ 1 – 2 + a + 6 = 0

a + 5 = 0

a = – 5

Question 6.

If the product of the roots of 4x^{3} + 16x^{2} – 9x – a = 0, is 9, then find ‘a’. [AP – Mar. 19, 17; TS – Mar. 16; May 13, 12, 08]

Solution:

Given equation is 4x^{3} + 16x^{2} – 9x – a = 0

Comparing this equation with ax^{3} + bx^{2} + cx + d = 0

we get, a = 4, b = 16, c = – 9, d = – a

Given that, the product of the roots = 9

s_{3} = 9

⇒ \(\frac{-\mathrm{d}}{\mathrm{a}}\) = 9

⇒ \(\frac{-(-a)}{4}\) = 9

⇒ a = 36.

Question 7.

If α, β and 1 are the roots of x^{3} – 2x^{2} – 5x + 6 = 0, then find α and β. [AP – May, Mar. 2016; May 09, March 08]

Solution:

Given equation is x^{3} – 2x^{2} – 5x + 6 = 0

Comparing this equation with ax^{3} + bx^{2} + cx + d = 0

we get a = 1, b = – 2, c = – 5, d = 6

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

then s_{1} = α + β + 1

= \(\frac{-b}{a}=\frac{-(-2)}{1}\) = 2

⇒ α + β = 1 …………(i)

s_{3} = αβ . 1 = \(\frac{-d}{a}\)

⇒ αβ = \(\frac{-6}{1}\) =

⇒ αβ = – 6

(α – β)^{2} = (α + β)^{2} – 4αβ

= 1 – 4 ( -6)

= 1 + 24 = 25

⇒ α – β = 5 ……………(2)

Solve (1) and (2);

Question 8.

If α, β and γ are the roots of x^{3} + 2x^{2} + 3x – 4 = 0, then find α^{2}β^{2}. [May ’07]

Solution:

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

Since α, β and γ are the roots of x^{3} + 2x^{2} + 3x – 4 = 0 then

α + β + γ = \(\frac{-(-2)}{1}\) = 2

αβ + βγ + γα = \(\frac{3}{1}\) = 3

αβγ = \(\frac{-(-4)}{1}\) = 4

Σα^{2}β^{2} = α^{2}β^{2} + β^{2}γ^{2} + γ^{2}α^{2}

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

= 3^{2} – 2 . 4(2)

= 9 – 16 = – 7.

Question 9.

If α, β and γ are the roots of 4x^{3} – 6x^{2} + 7x + 3 = 0, then find the value of αβ + βγ + γα. [TS – Mar. 2019]

Solution:

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

Comparing this equation with ax^{3} + bx^{2} + cx + d = 0

where a = 4; b = – 6; c = 7; d = 3

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

αβ + βγ + γα = S_{2} = \(\frac{c}{a}=\frac{7}{4}\).

Question 10.

Find the relations between the roots and the coefficients of the cubic equation 3x^{3} – 10x^{2} + 7x + 10 = 0.

Solution:

Given cubic equation is 3x^{3} – 10x^{2} + 7x + 10 = 0

Comparing this equation with

ax^{3} + bx^{2} + cx + d = 0, we get

a = 3, b = – 10, c = 7, d = 10

Let α, β, γ be the roots of given equation

s_{1} = \(\frac{-b}{a}\)

α + β + γ = \(\frac{-(-10)}{3}=\frac{10}{3}\)

s_{2} = αβ + βγ + γα

= \(\frac{c}{a}=\frac{7}{3}\)

s_{3} = αβγ

= \(\frac{-\mathrm{d}}{\mathrm{a}}=\frac{-10}{3}\).

Question 11.

Write down the relations between the roots and the coefficients of the biquadratic equation x^{4} – 2x^{3} + 4x^{2} + 6x – 21 = 0.

Solution:

Given biquadratic equation is

x^{4} – 2x^{3} + 4x^{2} + 6x – 21 = 0 ……………… (1)

Comparing this equation with

ax^{4} + bx^{3} + cx^{2} + dx + e = 0,

we get a = 1, b = – 2; c = 4; d = 6; e = – 21

Let α, β, γ, δ are the roots of equation (1) then

i) s_{1} = α + β + γ + δ

= Σα = \(\frac{-b}{a}=\frac{-(-2)}{1}\) = 2

ii) s_{2} = Σαβ

= \(\frac{c}{a}=\frac{4}{1}\) = 4

iii) s3 = Σαβγ

= \(\frac{-d}{a}=\frac{-6}{1}\) = – 6

iv) s = Σαβγδ

= \(\frac{\mathrm{e}}{\mathrm{a}}=\frac{-21}{1}\) = – 21

Question 12.

If 1, 2, 3 and 4 are the roots of x^{4} + ax^{3}+ bx^{2} + cx + d = 0, then find the values of a, b, c and d. [AP – May 2015]

Solution:

Given that the roots of the polynomial equation are 1, 2, 3 and 4.

Then(x – 1) (x – 2) (x – 3) (x – 4) = 0

(x^{2} – 3x + 2) (x^{2} – 7x + 12) = 0

x^{4} – 7x^{3} + 12x^{2} – 3x^{3} + 21x^{2} – 36x + 2x^{2} – 14x + 24 = 0

x^{4} – 10x^{3} + 35x^{2} – 50x + 24 = 0

Now, comparing this equation with

x^{4} + ax^{3} + bx^{2} + cx + d = 0

we get a = – 10; b = 35; c = – 50; d = 24.

Question 13.

If a, b, c are the roots of x^{3} – px^{2} + qx – r = 0 and r ≠ 0, then find \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\) in terms of p, q, r.

Solution:

Given equation is x^{3} – px^{2} + qx – r = 0

Since a, b and c are the roots of the equation x^{3} – px^{2} + qx – r = 0 then

s_{1} = a + b + c

= \(\frac{-(-p)}{1}\) = p;

s_{2} = ab + bc + ca

= \(\frac{q}{1}\) = q;

s_{3} = abc

= \(\frac{-(-r)}{1}\) = r

\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{b^2 c^2+a^2 c^2+a^2 b^2}{a^2 b^2 c^2}\)

= \(\frac{(a b+b c+c a)^2-2 a b c(a+b+c)}{(a b c)^2}\)

= \(\frac{q^2-2 p \cdot r}{r^2}=\frac{q^2-2 p r}{r^2}\).

Question 14.

Find the sum of the squares and the sum of the cubes of the roots of the equation x^{3} – px^{2} + qx – r = o in terms of p, q, r.

Solution:

Given equation is x^{3} – px^{2} + qx – r = 0

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

s_{1} = α + β + γ

= \(\frac{-\mathrm{b}}{\mathrm{a}}=\frac{-(-\dot{\mathrm{p}})}{1}\) = p;

s_{2} = αβ + βγ + γα

= \(\frac{q}{1}\) = q;

s_{3} = αβγ

= \(\frac{-(-r)}{1}\) = r

i) The sum of the squares of the roots of the equation = α^{2} + β^{2} + γ^{2}

= (α + β + γ)^{2} – 2(αβ + βγ + γα)

= p^{2} – 2q

ii) The sum of the cubes of the roots is α^{3} + β^{3} + γ^{3} = (α + β + γ)

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

= p(p^{2} – 2q – q) + 3r

= p^{3} – 3pq + 3r.

Question 15.

Let α, β, γ be the roots of x^{3} + px^{2} + qx + r = 0. Then find Σα^{3}. [March ’03]

Solution:

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

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

s_{1} = α + β + γ

= \(\frac{-\mathrm{p}}{1}\) = – p;

s_{2} = αβ + βγ + γα

= \(\frac{q}{1}\) = q;

s_{3} = αβγ

= \(\frac{-r}{1}\) = – r

Σα^{3} = α^{3} + β^{3} + γ^{3}

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

= (- p) ((p^{2} – 2q) – (q)) + 3 (- r)

= – p (p^{2} – 3q) – 3r

= – p^{3} + 3pq – 3r

Question 16.

Find s_{1}, s_{2}, s_{3} and s_{4} for the equation x^{4} – 16x^{3} + 86x^{2} – 176x + 105 = 0.

Solution:

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

Comparing this equation with ax^{4} + bx^{3} + cx^{2} + dx + e = 0

we get a = 1; b = – 16; c = 86; d = – 176; e = 105

Now, s_{1} = \(=\frac{-b}{a}=\frac{-(-16)}{1}\) = 16;

s_{2} = \(\frac{c}{\mathrm{a}}=\frac{86}{1}\) = 86;

s_{3} = \(\frac{-\mathrm{d}}{\mathrm{a}}=\frac{-(-176)}{1}\) = 176;

s_{4} = \(\frac{\mathrm{e}}{\mathrm{a}}=\frac{105}{1}\) = 105

Question 17.

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

Solution:

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

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

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

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

Question 18.

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:

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

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

Question 19.

Find the polynomial equation whose roots are the reciprocals of the roots of x^{4} – 3x^{3} + 7x^{2} + 5x – 2 = 0. [March ’11] [TS – Mar. 2015]

Solution:

Let f(x) = x^{4} – 3x^{3} + 7x^{2} + 5x – 2 = 0

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

⇒ \(\frac{1}{x^4}-\frac{3}{x^3}+\frac{7}{x^2}+\frac{5}{x}-2\) = 0

⇒ 1 – 3x + 7x^{2} + 5x^{3} – 2x^{4} = 0

⇒ 2x^{4} – 5x^{3} – 7x^{2} + 3x – 1 = 0

Question 20.

Form the polynomial equation whose roots are the squares of the roots of x^{3} + 3x^{2} – 7x + 6 = 0. [May ’02]

Solution:

Let f(x) = x^{3} + 3x^{2} – 7x + 6 = 0

The required equation is f(√x) = 0

(√x)^{3} + 3 (√x)^{2} – 7√x +6 = 0

⇒ x√x + 3x – 7√x + 6 = 0

⇒ 3x + 6 = – √x (x – 7)

Squaring on both sides

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

= x^{3} + 49x – 14x^{2}

x^{3} – 23x^{2} + 13x – 36 = 0.

Question 21.

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

Solution:

Let f(x) = x^{3} + 3x^{2} + 2 = 0

The required equation is f(\(\sqrt[3]{x}\))= 0

\((\sqrt[3]{x})^3+3(\sqrt[3]{x})^2\) + 2 = 0

⇒ x + 3 . x^{2/3} + 2 = 0

x + 2 = – 3x^{2/3}

Cubing on both sides (x + 2)^{3} = – 27x^{2}

⇒ x^{3} + 8 + 6x^{2} + 12x + 8 = – 27x^{2}

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

Question 22.

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

i) Σα^{2}

ii) Σα^{3}

iii) Σ \(\frac{1}{\alpha}\)

Solution:

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

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

s_{1} = α + β + γ = \(\frac{-\mathrm{p}}{1}\) = – p;

s_{2} = αβ + βγ + γα = \(\frac{q}{1}\) = q;

s_{3} = αβγ = \(\frac{-r}{1}\) = – r.

i) Σα^{2} = α^{2} + β^{2} + γ^{2}

= (α + β + γ)^{2} – 2(αβ + βγ + γα)

= (- p)^{2} – 2(q)

= p^{2} – 2q

ii) Σα^{3} = α^{3} + β^{3} + γ^{3}

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

= (- p) ((p^{2} – 2q) – (+ q)) + 3(- r)

= – p (p^{2} – 2q – q) – 3r

= – p^{3} + 3pq – 3r

iii) \(\Sigma \frac{1}{\alpha}=\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\)

= \(\frac{\beta \gamma+\alpha \gamma+\alpha \beta}{\alpha \beta \gamma}=\frac{q}{-r}=\frac{-q}{r}\)

Question 23.

Form the polynomial equation of the lowest degree with roots as 0, 0, 2, 2, – 2, – 2.

Solution:

The polynomial equation whose roots are 0, 0, 2, 2, – 2, – 2 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 – 4x) (x^{2} + 4 + 4x) = 0

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

⇒ x^{2} (x^{4} + 16 + 8x^{2} – 16x^{2}) = 0

⇒ x^{2} (x^{4} – 8x^{2} + 16) = 0

⇒ x^{6} – 8x^{4} + 16x^{2} = 0.

Question 24.

Form the monic polynomial equation of degree 3 whose roots are 2, 3 and 6. [May ’14, March ’02]

Solution:

x^{3} – 11x^{2} + 36x – 36 = 0

Question 25.

Form the monic polynomial equation of degree 4 whose roots are 4 + √3, 4 – √3, 2 + i and 2 – i.

Solution:

x^{4} – 12x^{3} + 50x^{2} – 92x + 65 = 0

Question 26.

Find s_{1}, s_{2}, s_{3} and s_{4} for the equation 8x^{4} – 2x^{3} – 27x^{2} – 6x + 9 = 0.

Solution:

s_{1} = \(\frac{1}{4}\),

s_{2} = \(\frac{-27}{8}\),

s_{3} = \(\frac{3}{4}\),

s_{4} = \(\frac{9}{8}\)

Question 27.

Find the algebraic equation whose roots are 3 times the roots of x^{3} + 2x^{2} – 4x + 1 = 0.

Solution:

x^{3} + 6x^{2} – 36x + 27 = 0

Question 28.

Find an algebraic equation of degree 4 whose roots are 3 times the roots of the equation 6x^{4} – 7x^{3} + 8x^{2} – 7x + 2 = 0. [March ’09]

Solution:

6x^{4} – 21x^{3} + 72x^{2} – 189x + 162 = 0.

Question 29.

Find the transformed equation whose roots are the negatives of the roots of x^{4} + 5x^{3} + 11x + 3 = 0. [AP – Mar. 2015]

Solution:

x^{4} – 5x^{3} – 11x + 3 = 0

Question 30.

Find the polynomial equation of degree 4 whose roots are the negatives of the roots of x^{4} – 6x^{3} + 7x^{2} – 2x + 1 = 0.

Solution:

x^{4} + 6x^{3} + 7x^{2} + 2x + 1 = 0

Question 31.

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:

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

Question 32.

Find the polynomial equation whose roots are the reciprocals of the roots of the equation x^{4} + 3x^{3} – 6x^{2} + 2x – 4 = 0.

Solution:

4x^{4} – 2x^{3} + 6x^{2} – 3x – 1 = 0

Question 33.

Find the polynomial equation whose roots are the squares of the roots of x^{4} + x^{3} + 2x^{2} + x + 1 = 0.

Solution:

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

Question 34.

Find the polynomial equation whose roots are the squares of the roots of x^{3} – x^{2} + 8x – 6 = 0.

Solution:

x^{3} + 15x^{2} + 52x – 36 = 0