Questions - Answers
1. Find the polynomial equation whose roots are the reciprocals of the roots of
x4-3x3+7x2+5x-2 = 0.
Sol: Given equation is f(x) = x4-3x3+7x2+5x-2 = 0
Required equation is x4f(
) = 0
x4 [()4 - 3()3 + 7()2 + 5() - 2] = 0
[1-3x + 7x2 + 5x3 - 2x4] = 0
or 2x4 - 5x3 - 7x2 + 3x - 1 = 0.
2. If the roots of x3+3px2+3qx+r = 0 are in Geometric Progression, find the condition.
Sol: The roots are in G.p.
Let the roots be , a, ar.
given (a/r) (a) (ar) = - r
a3 = - r
a = (- r)1/3
'a' is root of the given equation
a3+3pa2+3qa+r = 0
[(-r)1/3]3 + 3p((-r)1/3)2 + 3q(-r)1/3 + r = 0
p3r = q is the required condition.
3. If α, β, γ are roots of the equations x3-10x2+6x-8 = 0. Find α2 + β2 + γ2.
Sol: From the given equation,
α+ β+ γ = 10
αβ+ βγ + γα = 6
αβγ = 8
Now, α2+ β2+ γ2 = (α+ β+ γ)2 - 2 (αβ+ βγ + γα )
= (10)2 - 2(6)
= 100 - 12 = 88.
4. If α, β, l are the roots of x3-2x2-5x+6 = 0 then find α and β.
Sol: α, β and l are the roots of
x3-2x2-5x+6 = 0
α + β + 1 = 2 α+ β = 1
(α) (β) (1) = -6 ⇒ αβ = -6
(α - β)2 = (α+β)2 - 4β
= 1 + 24
= 25
α - β = 5 and α + β = 1
gives α =3 and β = -2 or
β = 3 and α = -2
5. If 1, -2 and 3 are the roots of x3-2x2+ax+6 = 0 then find a.
Sol: 1, -2 and 3 are roots of x3-2x2+ax+6 = 0
we know that if α, β, γ are the roots then αβ+ βγ + γα = a
⇒ (1) (-2) + (-2) (3) + (3) (1) = a
⇒ a = -5
6. Find the algebraic equation whose roots are 3 times the roots of x3+2x2-4x+1 = 0
Sol: Let the given equation be
f(x) = x3+2x2-4x+1 = 0.
The algebraic equation whose roots are 3 times that of f(x) = 0 is f(
) = 0
⇒ (x/3)3 + 2(x/3)2 -4 (x/3) + 1 = 0
⇒ x3 + 6x2 - 36x + 27 = 0
7. From the polynomial equation with rational co-efficients whose roots are 1 + 5i, 5 - i.
Sol: Since the 1+5i, 5-i are roots of the equation f(x) = 0, 1-5i, 5+i are also the roots of it.
.'. f(x) = (x - (1 + 5i)) (x - (5 - i)) (x - (1 - 5i)) (x - (5 + i)) = 0.
(x - 1 - 5i) (x - 1 + 5i) (x - 5+ i) (x - 5 - i) = 0
((x - 1)2 - (5i)2) ((x - 5)2 - (i)2) = 0
(x2 - 2x + 1 + 25) (x2 - 10x + 25 + 1) = 0
(x2 - 2x + 26) (x2 - 10x + 26) = 0
x4 - 12x3 + 72x2 - 312x + 676 = 0 is the required equation.
7 Marks questions
1. Solve x4 + 4x3 - 2x2 - 12x + 9 = 0 given that it has two pairs of equal roots.
Sol: Given equation is x4 + 4x3 - 2x2 - 12x + 9 = 0
Let the roots be α, α, β, β.
Sum of roots 2(α+ β) = -4 => α + β = -2.
Product of roots α2β2 = 9
⇒ αβ = ± 3.
Let αβ = -3 If αβ = 3
(α- β)2 = (α + β)2 - 4αβ (α- β)2 = (α + β)2 - 4αβ
= (-2)2 - 4(-3) = (-2)2 - 4(3)
= 4 + 12 = 4 - 12
= 16. = -8 < 0 not possible.
.'. α - β = 4
α + β = -2
----------------------------
2α = 2
α = 1 & β = -3.
.'. The roots are 1, 1, -3, -3.
2. Solve x4 + x3 - 16x2 - 4x + 48 = 0, given that the product of two of the roots is 6.
Sol: Let the roots of the given equation be α, β, γ and δ
Let αβ = 6 ....... (1)
Now we have α + β + γ + δ = -1 ......... (2)
αβ + αγ + αδ + βγ + βδ + γδ = -16 ......... (3)
αβγ + βγδ + γδα + δαβ = 4 ....... (4)
αβγδ = 48 ............ (5)
(1) & (5) γδ = 8 ----- (6)
Sub. αβ = 6 and γδ = 8 in (3)
We get (α + β) (γ + δ) = -16-8-6
(α + β) (γ + δ) = -30 ---- (7)
(1) & (7) ⇒
[(α + β) - (γ + δ)]2 = [(α + β) + (γ + δ)]2 - 4(α + β) (γ + δ)
= (-1)2 - 4 (-30)
= 1 + 120
= 121
(α + β) - (γ + δ) = ± 11
(α + β) - (γ + δ) = 11 & (α + β) + (γ + δ) = -1
⇒ 2(α + β) = 10 ⇒ α + β = 5 & γ + δ = -6
(α - β)2 = (α + β)2 - 4αβ (γ - δ)2 = (γ + δ)2 - 4γδ
= (5)2 - 4(6) = (-6)2 - 4(8)
= 25-24 = 1 ⇒ α - β = ± 1 = 36 - 32 = 4 ⇒ γ - δ = ± 2
Solving α + β = 5 Solving γ + δ = - 6
and α + β = ± 1 and γ - δ = ± 2
We get we get
α = 3 & β = 2 γ = -2 & δ = -4
or α = 2, & β = 3 or γ = -4 & δ = -2
The roots of the given equation are -4, -2, 2, 3.
3. Solve 18x3 + 81x2 + 121x + 60 = 0 given that one root is equal to half the sum of the remaining roots.
Sol: Given equation is
18x3 + 81x2 + 121x + 60 = 0
Let the roots of (1) be α, β and γ. By given condition let α = 
⇒ 2α = β + γ - (2)
Now, we have α + β + γ = -9/2 - (3)
αβ + βγ + γα = 121/18 - (4)
αβγ = -10/3 - (5)
(2) and (3) gives α = -3/2 and β + γ = - 3.
Sub α = -3/2 in (5), βγ = 20/9 (β - γ)2 = (β + γ)2 - 4βγ = 9 - 80/9 = 1/9
β - γ = ± 1/3 & β + γ = -3
⇒ β = -4/3 and γ = -5/3
or γ = -4/3 and β = -5/3
.
4. Solve 6x3 - 11x2 + 6x-1 = 0 given that the roots are in H.P.
Sol: Let f(x) = 6x3 -11x2 + 6x-1
⇒ The roots of f(x) = 0 are in H.P.
⇒ The roots of f() = 0 are in A.P.
f() = 6()3 - 11()2 + 6() - 1 = 0
6 - 11x + 6x2 - x3 = 0
x3 - 6x2 + 11x - 6 = 0 ---(1)
Let the roots of (1) be a-d, a and a+d.
Now we have a - d + a + a + d = 6 -- (2)
⇒ 3a = 6
⇒ a = 2
(a - d) a + a(a + d) +(a - d) (a + d) = 11 ---- (3)
and (a - d) a (a + d) = 6 -- (4)
Sub a = 2 in (4) ⇒ (2 - d) 2 (2 + d) = 6
⇒ 4 - d2 = 3 => d2 = 1
⇒ d = ± 1
.'. The roots of (1) are 1, 2, 3
and hence the roots of the given equation are 1, 1/2, 1/3
5. Find the polynomial equation whose roots are the translates of those of the equation.
x4-5x3+7x2-17x+11 = 0 by -2.
Sol: Given equation is
f(x) = x4 - 5x3 + 7x2 - 17x + 11 = 0.
Now the required equation is f(x+2) = 0.
i.e., (x+2)4 - 5(x+2)3 + 7(x+2)2 - 17(x+2) + 11 = 0.
i.e, A0x4 + A1x3 + A2x2 + A3x + A4 = 0.
.'. The required equation is x4 + 3x3 + x2 - 17x - 19 = 0
6. Solve the equation 6x6 - 25x5 + 31x4 - 31x2 + 25x - 6 = 0.
Sol: Given equation is a reciprocal equation of class II and is of even degree and hence ± 1 are the roots of the given equation 6x6 - 25x5 + 31x4 - 31x2 + 25x - 6 = 0 --- (1)
(1) = (x2-1) (6x4 - 25x3 + 37x2 - 25x + 6) = 0
6x4 - 25x3 + 37x2 - 25x + 6 = 0 --- (2)
Dividing by x2
Sub. these values in (3)
6(t2-2) -25t + 37 = 0
⇒ 6t2 - 25t + 25 = 0
⇒ 6t2 - 15t - 10t + 25 = 0
⇒ 3t(2t-5) - 5(2t-5) = 0
⇒ (3t-5) (2t-5) = 0
⇒ 2x2-4x-x+2 = 0
⇒ 2x(x-2) -1 (x-2) = 0
⇒ (2x-1) (x-2) = 0 x = 1/2 or 2.
.'. The roots of the given equation (1) are
