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## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.2

Question 1.

If k is real, discuss the nature of the roots of the polynomial equation 2x^{2} + kx + k = 0, in terms of k.

Solution:

Δ = b² – 4ac

a = 2, b = k, c = k

Δ = k² – 4 × 2(k)

Δ = k² – 8k

when k < 0, the polynomial has real roots (Δ > 0)

If Δ = 0 k² – 8k = 0

k(k – 8) = 0

k = 0 or k = 8

When k = 0 or k = 8. The roots are real and equal.

When 0 < k < 8, (Δ < 0) the roots are imaginary.

when k > 8. The roots are real and distinct.

Question 2.

Find a polynomial equation of minimum degree with rational coefficients, having 2 + √3 i as a root.

Solution:

Given roots is (2 + √3 i)

The other root is (2 – √3 i), since the imaginary roots with real co-efficient occur as conjugate pairs.

x^{2} – x(S.O.R) + P.O.R = 0

⇒ x^{2} – x(4) + (4 + 3) = 0

⇒ x^{2} – 4x + 7 = 0

Question 3.

Find a polynomial equation of minimum degree with rational coefficients, having 2i + 3 as a root.

Solution:

Let the root be 3 + 2i

Another root be 3 – 2i

Sum of the roots = 3 + 2i + 3 – 2i = 6

Product of the roots = (3 + 2i) (3 – 2i) = 3² + 2² = 9 + 4 = 13

Required equation is x² – (SR)x + PR = 0

x² – 6x + 13 = 0

Question 4.

Find a polynomial equation of minimum degree with rational coefficients, having √5 – √3 as a root.

Solution:

The given one roots of the polynomial equation are (√5 – √3)

The other roots are (√5 + √3), (-√5 + √3) and (-√5 – √3).

The quadratic factor with roots (√5 – √3) and (√5 + √3) is

= x^{2} – x(S.O.R) + P.O.R

= x^{2} – x(2√5) + (√5 – √3) (√5 + √3)

= x^{2} – 2√5 x + 2

The other quadratic factors with roots (-√5 + √3) (-√5 – √3) is

= x^{2} – x (S.O.R) + P.O.R

= x^{2} – x (-2√5 ) + (5 – 3)

= x^{2} + 2√5x + 2

To rationalize the co-efficients with minimum degree

(x^{2} – 2√5 x + 2) (x^{2} + 2√5 x + 2) = 0

⇒ (x^{2} + 2)^{2} – (2√5 x)^{2} = 0

⇒ x^{4} + 4 + 4x^{2} – 20x^{2} = 0

⇒ x^{4} – 16x^{2} + 4 = 0

Question 5.

Prove that a straight line and parabola cannot intersect at more than two points.

Solution:

let be the equation of a straight line y = mx + c

Let be the equation of a parabola y² = 4ax

(mx + c)² = 4ax

m²x² + 2mcx + c² – 4ax = 0

m²x² + (2x)(mc – 2a) + c² = 0

Which is a quadratic equation.

It can not have more than two solutions.

### Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.2 Additional Problems

Question 1.

Find a polynomial equation of minimum degree with rational co-efficients having 1 – i as a root.

Solution:

Given root is 1 – i

The other root is 1 + i

Sum of the roots: 1 – i + 1 + i = 2

product of the roots: (1 – i) (1 + i) = (1)^{2} + (1)^{2} ⇒ 1 + 1 = 2

∴ The required polynomial equation of minimum degree with rational coefficients is

x^{2} – x (S.R.) + (P.R.) = 0

x^{2} – 2x + 2 = 0

Question 2.

Find a polynomial equation of minimum degree with rational coefficients having \(\sqrt{3}+\sqrt{7}\) as a root.

Solution:

The required polynomial equation of minimum degree

Question 3.

If the roots of the equation x^{3} + px^{2} + qx + r = 0 are in A.P then show that 2p^{3} – 9pq + 27 r = 0.

Solution:

Let the roots of the given equation is a – d, a, a + d

Question 4.

Solve 27x^{3} + 42x^{2} – 28x -8 = 0 given that its roots are in geometric progressive.

Solution:

Question 5.

Solve the equation 15x^{3} – 23x^{2} + 9x – 1 = 0. Where roots are in H.P.

Solution: