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

Question 1.

Solve the cubic equation: 2x^{3} – x^{2} – 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 9x^{3} – 36x^{2} + 44x – 16 = 0 if the roots form an arithmetic progression.

Solution:

The given equation is 9x^{3} – 36x^{2} + 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 3x^{3} – 26x^{2} + 52x – 24 = 0 if its roots form a geometric progression.

Solution:

The given equation is 3x^{3}– 26x^{2} + 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 2x^{3} – 6x^{2} + 3x + k = 0 if one of its roots is twice the sum of the other two roots.

Solution:

The given equation is 2x^{3} – 6x^{2} + 3x + k = 0

Question 5.

Find all zeros of the polynomial x^{6} – 3x^{5} – 5x^{4} + 22x^{3} – 39x^{2} – 39x + 135, if it is known that 1 + 2i and √3 are two of its zeros.

Solution:

The given equation is x^{6} – 3x^{5} – 5x^{4} + 22x^{3} – 39x^{2} – 39x + 135 = 0

The given roots are 1 + 2i, √3

The other roots are 1 – 2i, -√3

The factors are

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

= (x^{2} – 2x + 5) (x^{2} – 3)

= x^{4} – 3x^{2} – 2x^{3} + 6x + 5x^{2} – 15

= x^{4} – 2x^{3} + 2x^{2} + 6x – 15

To find this roots,

Question 6.

Solve the cubic equations:

(i) 2x^{3} – 9x^{2} + 10x = 3

(ii) 8x^{3} – 2x^{2} – 7x + 3 = 0

Solution:

(i) Given equation is 2x^{3} – 9x^{2} + 10x = 3

Sum of the co-efficients = 0

(x – 1) is a factor.

The other factor is 2x^{2} – 7x + 3

2x^{2} – 7x + 3 = 0

(x – 3)(2x – 1) = 0

The roots are 1, 3, \(\frac { 1 }{ 2 }\)

(ii) Given equation is 8x^{3} – 2x^{2} – 7x + 3 = 0

Sum of odd co-efficients = Sum of even co-efficients

(x + 1) is a factor.

The other factor is 8x^{2} – 10x + 3

8x^{2} – 10x + 3 = 0

(4x – 3) (2x – 1) = 0

The roots are \(\frac{3}{4}, \frac{1}{2},-1\)

Question 7.

Solve the equation: x^{4} – 14x^{2} + 45 = 0.

Solution:

The given equation is x^{4} – 14x^{2} + 45 = 0

Let x^{2} = y

y^{2} – 14y + 45 = 0

(y – 9)(y – 5) = 0

y = 9, 5

x^{2} = 9, x^{2} = 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 x^{3} + 2x^{2} + 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 x^{3} – 3x^{2} – 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 x^{3} – 5x^{2} + 9x – 5 = 0, If 2 + i is a root.

Solution: