Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 3 Theory of Equations Ex 3.2 Textbook Questions and Answers, Notes.

## 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:

The given quadratic equation is 2x^{2} + kx + k = 0

a = 2, b = k, c = k

∆ = b^{2} – 4ac = k^{2} – 4(2) k = k^{2} – 8k

(i) If the roots are equal

k^{2} – 8k = 0

⇒ k(k – 8) = 0

⇒ k = 0, k = 8

(ii) If the roots are real

k^{2} – 8k > 0

k(k – 8) > 0

k ∈ (-∞, 0) ∪ (8, ∞)

(iii) If this roots are imaginary

k^{2} – 8k < 0

⇒ k ∈ (0, 8)

Question 2.

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

Solution:

Let the root be 2 + i √3

Another root be 2 – i √3

Sum of the roots = 2 + i √3 + 2 – i √3 = 4

Product of the roots = (2 + i √3) (2 – i √3) = 2² + √3² = 4 + 3 = 7

x² – (SR)x + PR = 0

x² – 4x + 7 = 0

Question 3.

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

Solution:

Given roots is (3 + 2i), the other root is (3 – 2i);

Since imaginary roots occur in with real co-efficient occurring conjugate pairs.

x^{2} – x(S.O.R) + P.O.R = 0

⇒ x^{2} – x(6) + (9 + 4) = 0

⇒ x^{2} – 6x + 13 = 0

Question 4.

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

Solution:

Let the root be √5 – √3,

Another root is √5 + √3

Sum of the roots = √5 – √3 + √5 + √3 = 2√5

Product of roots = (√5 – √3) (√5 + √3)

√5² – √3² = 5 – 3 = 2

x² – (SR)x + PR = 0

x² – 2√5 x + 2 = 0 which is not rational co-efficient.

to make rational co-efficient

(x² + 2√5 x + 2) (x² + 2 + 2√5 x) = 0

(x² + 2)² – (2√5x)² = 0

x^{4} + 4 + 4x² – 20x² = 0

⇒ x^{4} – 16x² + 4 = 0 is a rational co-efficient polynomial equation.

Question 5.

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

Solution:

Let the standard equation of parabola y^{2} = 4ax …..(1)

Equation of line be y = mx + c …(2)

Solving (1) & (2)

(mx + c)^{2} = 4ax

⇒ mx^{2} + 2mcx + c^{2} – 4ax = 0

⇒ mx^{2} + 2x(mc – 2a) + c^{2} = 0

This equation can not have more than two solutions and

hence a line and parabola cannot intersect at more than two points.