Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.3

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.3

Question 1.
Solve the cubic equation: 2x³ – x² – 18x + 9 = 0 if sum of two of its roots vanishes.
Solution:
2x³ – x² – 18x + 9 = 0
Let the roots be α, β and γ
given α + β = 0
α + β + γ = \(\frac{1}{2}\)
γ = \(\frac{1}{2}\) (∵ α + β = 0)
x² + (α + β)x + αβ = 0, γ = \(\frac{1}{2}\)
x² – 0x + p = 0, (x – \(\frac{1}{2}\)) = 0
(x² + p) = 0, 2x – 1 = 0
∴ (x² + p) (2x – 1) = 2x³ – x² – 18x + 9
2x³ – x² + 2px – p = 2x³ – x² – 18x + 9
-p = 9
p = -9
x² – 9 = 0
x² = 9
x = ± 3
The roots are 3, -3, and \(\frac{1}{2}\)

Question 2.
Solve the equation 9x³ – 36x² + 44x – 16 = 0 if the roots form an arithmetic progression.
Solution:
The given equation is 9x³ – 36x² + 44x – 16 = 0
\(x^{3}-4 x^{2}+\frac{44}{9} x-\frac{16}{9}=0\)
Let the roots be α – d, α, α + d
As they are in A.P

Question 3.
Solve the equation 3x³ – 26x² + 52x – 24 = 0 if its roots form a geometric progression.
Solution:
The given equation is 3x³– 26x² + 52x – 24 = 0
\(x^{3}-\frac{26}{3} x^{2}+\frac{52}{3} x-8=0\)
Given that the root are GP

Question 4.
Determine ft and solve the equation 2x³ – 6x² + 3x + k = 0 if one of its roots is twice the sum of the other two roots.
Solution:
The given equation is 2x³ – 6x² + 3x + k = 0

Question 5.
Find all zeros of the polynomial x⁶ – 3x⁵ – 5x⁴ + 22x³ – 39x² – 39x + 135, if it is known that 1 + 2i and √3 are two of its zeros.
Solution:
The given equation is x⁶ – 3x⁵ – 5x⁴ + 22x³ – 39x² – 39x + 135 = 0
The given roots are 1 + 2i, √3
The other roots are 1 – 2i, -√3
The factors are
= {x² – x(2) + (1 + 4)}{(x + √3)(x – √3)}
= (x² – 2x + 5) (x² – 3)
= x⁴ – 3x² – 2x³ + 6x + 5x² – 15
= x⁴ – 2x³ + 2x² + 6x – 15
To find this roots,

Question 6.
Solve the cubic equations:
(i) 2x³ – 9x² + 10x = 3
(ii) 8x³ – 2x² – 7x + 3 = 0
Solution:
(i) Given equation is 2x³ – 9x² + 10x = 3
Sum of the co-efficients = 0
(x – 1) is a factor.
The other factor is 2x² – 7x + 3
2x² – 7x + 3 = 0
(x – 3)(2x – 1) = 0
The roots are 1, 3, \(\frac { 1 }{ 2 }\)

(ii) Given equation is 8x³ – 2x² – 7x + 3 = 0
Sum of odd co-efficients = Sum of even co-efficients
(x + 1) is a factor.
The other factor is 8x² – 10x + 3
8x² – 10x + 3 = 0
(4x – 3) (2x – 1) = 0
The roots are \(\frac{3}{4}, \frac{1}{2},-1\)

Question 7.
Solve the equation: x⁴ – 14x² + 45 = 0.
Solution:
The given equation is x⁴ – 14x² + 45 = 0
Let x² = y
y² – 14y + 45 = 0
(y – 9)(y – 5) = 0
y = 9, 5
x² = 9, x² = 5
x = ± 3, x = ± √5
The roots are ± 3, ± √5

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.3 Additional Problems

Question 1.
If one root of x³ + 2x² + 3x + k = 0 is sum of the other two roots then find the value of k.
Solution:
Let α, β, γ be the roots of given equation.
But, α = β + γ
α + β + γ = -2 … (1) ⇒ 2α = -2 ⇒ α = -1
αβ + βγ + γα = 3 … (2) This gives β + γ = – 1
αβγ = -k …(3)

Question 2.
If sum of the roots of the equation x³ – 3x² – 16x + k = 0 is zero then find the value of k.
Solution:
Let α, β, γ be the roots of the given equation.
But, α + β = 0

Question 3.
Find all zeros of the polynomial x³ – 5x² + 9x – 5 = 0, If 2 + i is a root.
Solution: