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

## 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 2x2 + 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.
x2 – x(S.O.R) + P.O.R = 0
⇒ x2 – x(4) + (4 + 3) = 0
⇒ x2 – 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
= x2 – x(S.O.R) + P.O.R
= x2 – x(2√5) + (√5 – √3) (√5 + √3)
= x2 – 2√5 x + 2
The other quadratic factors with roots (-√5 + √3) (-√5 – √3) is
= x2 – x (S.O.R) + P.O.R
= x2 – x (-2√5 ) + (5 – 3)
= x2 + 2√5x + 2
To rationalize the co-efficients with minimum degree
(x2 – 2√5 x + 2) (x2 + 2√5 x + 2) = 0
⇒ (x2 + 2)2 – (2√5 x)2 = 0
⇒ x4 + 4 + 4x2 – 20x2 = 0
⇒ x4 – 16x2 + 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
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
x2 – x (S.R.) + (P.R.) = 0
x2 – 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 x3 + px2 + qx + r = 0 are in A.P then show that 2p3 – 9pq + 27 r = 0.
Solution:
Let the roots of the given equation is a – d, a, a + d Question 4.
Solve 27x3 + 42x2 – 28x -8 = 0 given that its roots are in geometric progressive.
Solution: Question 5.
Solve the equation 15x3 – 23x2 + 9x – 1 = 0. Where roots are in H.P.
Solution:  