TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Students must practice these TS Inter 2nd Year Maths 2A Important Questions Chapter 4 Theory of Equations to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2A Theory of Equations Important Questions

Question 1.
Form the monic polynomial eduation of degree 3 whose roots are 2, 3 and 6. Given roots of required polynomial are
2, 3 and 6.
Solution:
We know that monic polynomial those roots of α, β and γ is (x -α) (x -β) (x -γ) = 0.
∴ Equation of required monk polynomial is
(x – 2)(x – 3)(x – 6) = 0
on simplification, it becomes
x³ – 11x² + 36x – 36 = 0.

Question 2.
Find the relations between the roots and the coefficients of the cubic equation
3x² – 10x² + 7x + 10 = 6.
Solution:
Given cubic equation is
3x² – 10x² + 7x + 10 = 0
On dividing the equation by ‘3’ we get,

Question 3.
Write down the relations between the roots and the coefficients of the biquadratic equation :
x⁴– 2x³ + 4x² + 6x – 21 = 0.
Solution:
Given equation is x⁴– 2x³ + 4x² + 6x – 21 = O
On comparing (1) with

Question 4.
If 1,2,3 and 4 are the roots of x⁴+ax³+bx²+ cx + d = 0, then find the values of a, b, c and d.
Solution:
Given roots of the given polynomal equation are 1, 2, 3 and 4.
∴ x⁴+ax³+bx²+ cx + d ≡ (x-1)(x-2) (x – 3) (x – 4)
= x⁴ – 10x³ 35x² – 50x + 24
On comparing coellicients of like powers of we get,
a = – 10, b = 35, c = – 50, d = 24.

Question 5.
If a,b and c are the roots of x³-px²+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 a ,b and c are the roots of

Question 6.
Find the sum of the squares and the sum of the cubes of the roots of the equation
x³ – px² + qx – r = 0 in terms of p, q, r.
Solution:
Let α, β and γ be roots of the given equation x³ – px² + qx – r = 0 ……………..(1)

Question 7.
Obtain the monic cubic equation, whose roots are the squares of the roots of the equation x³ + p₁x³ + p₂x + p₃ = 0.
Solution:
Let α, β and γ be the roots of the given equation x³ + p₁x³ + p₂x + p₃ = 0 ………………… (1)

Question 8.
Let α, β and γ be the roots of x³+px²+qx+r = 0. Then find
(i) Σα²
(ii) \(\Sigma \frac{1}{\alpha}\) ,If α, β,γ are non-zero
(iii) Σα³
(iv) Σβ²γ²
(v) (α+β)(β+γ)(γ+α)
Solution:
Let α, β, and γ be the roots of equation  x³+px²+qx+r = 0 ……………….(1)

Question 9.
If α, β, γ are the roots of x³ + ax² + bx + c = 0, then find ∑α²β + ∑αβ².
Solution:
Given α, β and γ  are the roots of

Question 10.
If α, β, γ are the roots of x³ + px² + qx + r = 0, then form the monic cubic equation whose rools are
α(β+ γ),β(y+α), y(α+ β).
Solution:
Given a, and y are the roots of

Question 11.
Solve x³ – 3x² – 16x + 48 = 0.
Solution:
Let f(x) = x³ – 3x² – 16x + 48
By inspection, we see that
f(3)= 27- 27 – 48+48 = 0
Hence 3 is a root of f(x) = 0
Now we divide 1(x) by (x-3), using synthetic division.

Thus the quotient is (x² – 16) and the remainder is 0.
Therefore f(x) (x – 3) (x² – 16)
= (x – 3) (x – 4) (x 4)
Hence 3, – 4, 4 are the roots of the given equation.

Question 12.
Find the roots of x⁴ – 16x³ + 86x² – 176x + 105 = 0.
Solution:
Let f(x) = x⁴ – 16x³ + 86x² – 176x + 105
By inspection we see that,
f(1)= 1- 16+86 – 176+ 105=0
Hence 1 is a root of 1(x) = 0
Now we divide f(x) by (x – 1), using synthetic division.

Therefore
x⁴ – 16x³ + 86x² – 176x + 105
(x – 1)(x³-15x² + 71x – 105) ……………… (1)
Let g(x) = x³ – 15x² + 71x – 105
By inspection g(3) = 0.
Hence 3 is a root of g(x) = 0.
Now we divide g(x) by (x -3),
using synthetic division.

Therefore g(x) (x – 3) (x² – 12x + 35) ……………. (2)
From (1) and (2),
f(x) (x – 1) (x – 3) (x² – 12x + 35)
= (x – 1) (x – 3) (x – 5) (x – 7)
Hence 1, 3, 5 and 7 are the roots of the given equation.
Now we solve equations when a relation between some of the roots is given.

Question 13.
Solve x³ – 7x² +36 = 0 given one root being twice the other.
Solution:
Let α, β, γ be the three roots of the given equation and β = 2a.
Now we have α+β+γ = 7
αβ + βγ + γα = 0
αβγ = – 36
On substituting β =2α in the above equations, we obtain
3α + γ = 7 …………………… (1)
2α²+3αγ = 0 ……………………. (2)
2α² γ = – 36 ……………………. (3)
On eliminating γ from (1) and (2), we have
2α² + 3α (7 – 3α) = 0
i.e., α² -3α = 0 or α(α -3) = 0
Therefore α = 0 or α = 3
Since α = 0 does not satisfy the given equation, we ignore this value.
Therefore α = 3 is a root of the given equation.
So, β = 6 (since β = 2α) and γ = – 2
Hence 3, 6, – 2 are the roots of the given equation.

Question 14.
Given that 2 is a root of x³ – 6x²+3x+10 = 0, find the other roots.
Solution:
Let f(x) x³ – 6x²+3x+10
Since 2 is a root of f(x) 0,
we divide f(x) by (x – 2)

Therefore
x³ – 6x² – 3x + 10 = (x – 2) (x² – 4x – 5)
= (x – 2) (x+1) (x-5)
Thus – 1, 2 and 5 are the roots of the given equation.

Question 15.
Given that two roots of 4x³ + 20x²– 23x + 6 = 0 are equal, find all the roots of the given equation.
Solution:
Let α, β, γ, and δ are the roots of
4x³ + 20x²– 23x + 6 = 0 ………………… (1)
Given that two roots of (1) are equal. Let β = α
Since α, β, γ are the roots of (1), we have

Therefore
x³ – 6x² – 3x + 10 = (x – 2) (x² – 4x – 5)
=(x – 2) (x+ 1) (x – 5)
Thus – 1, 2 and 5 are the roots of the given equation.
Therefore α = \(\frac{1}{2}\) or α = \(-\frac{23}{6}\)
On verification we get that α = \(\frac{1}{2}\) is a root of (1)
On substituting this value in (2), we get γ = – 6
Therefore \(\frac{1}{2}, \frac{1}{2},-6\) are the roots of (1).

Question 16.
Given that the sum of two roots of x⁴ – 2x³+ 4x² + 6x – 21 = 0 is zero. Find the roots of the equation.
Solution:
Let α, β, γ, and δ be the roots of (1), and α + β = 0 ……………. (1)
From the relation between the coefficients and the roots, we have
α + β + γ +δ = 2 so γ + δ = 2 ……………… (2)
Therefore the quadratic equation having roots α and β is x² – 0, x + αβ=0
The quadratic equation having roots γ + δ is x²-2x +q =0

Question 17.
Solve 4x³ – 24x²+ 23x+ 18=0 given that the roots of this equation are in arithmetic progression.
Solution:
Let a – d, a, a + d be the roots of the given equation. These are in A.P. From the relation between the coefficients and the roots, we have

Question 18.
Solve x³-7x²+14x-8=0 given that the roots are in geometric progression.
Solution:

Hence the roots of the given equation are 1, 2 and 4.

Question 19.
Solve x⁴-5x³+5x²+ 5x – 6 = 0 given that the product of two of its roots is 3.
Solution:
Let α , β, γ be the roots of the given equation.

we get γ = 3 and δ = 1 or γ = 1 and δ = 3
Hence the roots of the given equation are 2,- 1, 3 and 1.

Question 20.
Solve x⁴+4x³– 2x⁴– 12x+ 9 = 0 given that it has two pairs of equal roots.
Solution:
Let the roots of the given equation be

Therefore 1, 1, -3, -3 are the roots of the given equation.

Question 21.
Prove that the sum of any two of the roots of the equation x⁴ +px³ + qx² + rx + s = 0 is equal to the sum of the remaining two roots of the equation if p³ – 4pq + Sr = 0.
Solution:
Suppose that the sum of two of the roots of the given equation is equal to the sum of the remaining two roots.
Let α, β, γ, δ be roots of the given equation such that α + β = γ + δ


Take \(b=\frac{p}{2}\)
Then equations (1), (2) and (4) are satisfied.
In view of (5), equation (3) is also satisfied.
Hence (x² + bx + c) (x² + bx + d)
=x⁴+ 2bx³+(b²+c+d)x²+h(c +d)x+cd
= x⁴ +px³ +qx² +rx+s
Hence the roots of the given equation are α₁, β₁,γ₁ and δ₁, where α₁ and β₁ are the roots of the equations
x² + bx + c = 0 and γ₁ and δ₁ are those of the equation x² + bx+ d = 0
We have α₁ + β₁ = -b = γ₁ + δ₁

Question 22.
Form the monic polynomial equation of degree 4 whose roots are \(4+\sqrt{3}, 4-\sqrt{3}\)
Solution:
The required equation is
\(\{\mathrm{x}-(4+\sqrt{3})\}\{\mathrm{x}-(4-\sqrt{3})\}\)
{x – (2+i)} {x – (2 – i)) =0
i.e., (x² – 8x + 13) (x² – 4x + 5) = 0
i.e., x² – 12x³ + 50x – 92x + 65 = 0

Question 23.
Solve 6x⁴ -13x² – 35x² – x+3 = 0 given that one of its roots is \(2+\sqrt{3}\)
Solution:
Since \(2+\sqrt{3}\) is a root of the giver equation, by Theorem 4.3.9, \(2+\sqrt{3}\) is also a root of it. The quadratic factor corresponding to these two roots is x² – 4x + 1.
On dividing 6x⁴ -13x¹ – 35x² –  x + 3 by x² – 4x + 1 (by synthetic division) weet the quotient 6x² + 11x + 3.
Therefore 6x – 13x³ – 35x² – x + 3 =(x²– 4x+ 1) (6x²+ 11x+3)
Hence the other roots are obtained from 6x² + 11x + 3 = 0
On solving this equation, we get
\(\mathrm{x}=-\frac{1}{3} \text { or }-\frac{3}{2}\)
Thus the roots of the given equation are
\(-\frac{1}{3},-\frac{3}{2}, 2 \pm \sqrt{3}\)

Question 24.
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
By Theorem 4.4.1, the equation (- x) = 0 has the desired property.
We have
f(-x) = (-x)⁴ – 6(-x)³. 7(-x)² – 2(-x) + 1
= x⁴+6x³+7x²+2x+ 1
Hence x⁴ – 6x³ +7x² – 2x + 1 + 6x³ + 7x² + 2x+ 1 = 0 is the desired equation.

Question 25.
Find an algebraic equation of degree 4 whose roots are 3 times the roots of equation 6x⁴ – 7x³ + 8x² – 7x + 2 = 0.
Solution:
Let f(x) = 6x⁴ – 7x³ + 8x² – 7x + 2
By Theorem 4.4.3, the equation
\(f\left(\frac{x}{3}\right)=0\) has the desired properties. We have

Hence 6x⁴– 21x³+72x²-189x +162 = 0 is the desired equation.

Question 26.
Form the equation whose roots are m times the roots of the equation
\(x^3+\frac{1}{4} x^2-\frac{1}{16} x+\frac{1}{72}=0\) and deduce the case when m = 12.
Solution:
From the note 4.4.2 and note 4.4.4., it follows that

is a polynomial equation of degree 3, whose roots are m times those of the given equation. On taking m = 12, equation (1) reduces to

Question 27.
Find the polynomial equation of degree 5 whose roots are the translates of the roots of x⁵+4x³ -x²+11 = 0 by – 3.
Solution:
By Theorem 4.4.6, the equation
(x+3)⁵+4(x+3)³_(x+3)² +11 = 0 has the desired properties.
On simplifying the above equation, we get
x⁵ + 15x⁴ + 94x³ + 305x² + 507x + 353 = 0.
The transformed equation can also be object by synthetic division
Let f(x) =x⁵+4x³-x²+ 11
Suppose that f(x + 3) A₀x⁵ + A₁x⁴ + A₂x³ + A₃x² + A₄x + A₅
Then by note 4.4.8, the coefficients
A₀, A₁ , …………………………. A₅ can be obtained as follows

Therefore the roots of the equation
x⁵ + 15x⁴ + 94x³ + 305x² + 507x + 353 = 0 are the translates of the roots of the given equation by – 3.

Question 28.
Find the algebraic equation of degree 4 whose roots are the translates of the roots of 4x⁴ + 32x³ + 83x² + 76x + 21 = O by 2.
Solution:
By Theorem 4.4.6, the equation
4 (x – 2)⁴ + 32 (x – 2)³ + 83 (x – 2)⁴+76(x-2) + 21 =0 has desired properties.
On simplifying the above equation, we get 4x⁴– 13x²+90
Other mehod: The equation A₀x⁴ . A₁x³ +………… + A₄ = 0 whose coefficients are obtained by synthetic division as given below, has the desired properties

Hence the equation 4x⁴ – 13x² + 9 = 0 has the desired properties.

Question 29.
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
By theorem 4.4.9, the equation
\(x^4 f\left(\frac{1}{x}\right)=0\) has the desired properties.
Therefore
\(x^4\left[\frac{1}{x^4}+3 \frac{1}{x^3}-6 \frac{1}{x^2}+\frac{2}{x}-4\right]=0\)
i.e., 4x⁴ – 2x³ + 6x² – 3x – 1 = 0 is the required equation.

Question 30.
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.
Then as per the notation introduced in Note 4.413, we have

The equation x³+ 15x² + 52x -36 = 0 has the desired properties.

Question 31.
Show that 2x³+5x²+5x+2=0 is are reciprocal equation of class one.
Solution:
Let f(x) = 2x³+5x²+5x+2=0
Then 2 is the leading coefficient.

Therefore the given equation is a reciprocal equation of class one.

Question 32.
Solve the equation 4x³ -13x² – 13x + 4 = 0.
Solution:
The given equation is an odd degree reciprocal equation of class one.
By Note 4.4.24(1), – 1 is a root of this equation.
Therefore (x + 1) is a factor of
4x³ -13x² – 13x + 4
Hence on dividing this expression by(x + 1), we get
4x³ -13x² – 13x + 4 (x + 1)(4x² – 17x+ 4)
Now the roots of the equation
4x²– 17x+4=0 are \(\frac{1}{4}\) and 4.
Therefore -1,\(\frac{1}{4}\),4 are the roots of the given equation.

Question 33.
Solve the equation
– 5x⁴ + 9x³ – 9x² + 5x – 1 = 0.
Solution:
We observe that the given equation is an odd degree reciprocal equation of class two.
By Note 4.4.24(1), 1 is a root of this equation.
Therefore (x – 1) is a factor of x⁵ – 5x⁴ + 9x³– 9x² + 5x – 1.
On dividing this expression by (x – 1), we get
x⁴ – 4x³ + 5x² – 4x + 1 as the quotient.
Now we have to solve the equation
x⁴ – 4x + 5x² – 4x + 1 = 0
On dividing this equation by x², we get
x²– 4x+5 –\(\frac{4}{x}+\frac{1}{x^2}\) = 0

Question 34.
Solve the equation 6x⁴-35x³+62x²-35x+6=0.
Solution:
We observe that the given equation is an even degree reciprocal equation of class one.
On dividing both sides of the given equation by x², we get

Then the above equation reduces to
6 (y²-2) – 35y + 62 = 0
i.e., 6y² – 35y + 50 = 0
i.e., (2y – 5) (3y – 10) = 0
Hence the roots of

Question 35.
Solve the equation
6x⁶ – 25x⁵ + 31x⁴ -31x² + 25x – 6 = 0.
Solution:
We observe that the given equation is an even degree reciprocal equation of class two. By Note 4.4.24(2), + 1 and – 1 are the roots of this equation. Hence (x + 1) and (x – 1) are the factors of the given equation.
Let f(x) = 6x⁶ – 25x⁵ + 31x⁴ -31x² + 25x – 6 = 0
On dividing this expression by (x + 1) and then by (x – 1), we get
f(x) = (x² -1) (6x⁴ – 25x³ + 37x² – 25x + 6)
Now we have to solve the equation
6x⁴ – 25x³ + 37x² -25x + 6 = 0
On dividing both sides of this equation by x², we obtain