Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 3 Theory of Equations Ex 3.6 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.6

Question 1.

Discuss the maximum possible number of positive and negative roots of the polynomial equation 9x^{9} – 4x^{8} + 4x^{7} – 3x^{6} + 2x^{5} + x³ + 7x² + 7x + 2 = 0.

Solution:

P(x) = 9x^{9} – 4x^{8} + 4x^{7} – 3x^{6} + 2x^{5} + x^{3} + 7x^{2} + 7x + 2

The number of sign changes in P(x) is 4.

∴P(x) has at most 4 positive roots.

P(-x) = -9x^{9} – 4x^{8} – 4x^{7} – 3x^{6} – 2x^{5} – x^{3} + 7x^{2} – 7x + 2

The number of sign changes in P(-x) is 3.

P(x) has almost 3 negative roots.

Since the difference between the number of sign changes in co-efficient P(-x) and the number of negative roots of the polynomial P(x) is even.

The number of negative roots = at most 2.

Question 2.

Discuss the maximum possible number of positive and negative roots of the polynomial equations x² – 5x + 6 and x² – 5x + 16 . Also, draw a rough sketch of the graphs.

Solution:

y = x² – 5x + 6

x = 1, y = 1 – 5 + 6 = 2

x = 2, y = 4 – 10 + 6 = 0

x = 0, y = 6

x = 3, y = 9 – 15 + 6 = 0

x = -1, y = 1 + 5 + 6 = 12

x = 4, y = 16 – 20 + 6 = 2

(1, 2), (0, 6), (-1, 12)

P(x) = (x² + 5x + 6) (x² – 5x + 16)

= x^{4} – 5x³ + 16x² – 5x³ + 25x² – 80x + 6x² – 30x + 96 = 0

= x^{4} – 10x³ + 47x² – 110x + 96 = 0

It has two sign changes

∴ It has two positive real roots

P(-x) = x^{4} + 10x³ + 47x² + 110x + 96

It has no sign change, no negative real roots

y = x^{2} – 5x + 16

Question 3.

Show that the equation x^{9} – 5x^{5} + 4x^{4} + 2x² + 1 = 0 has atleast 6 imaginary solutions.

Solution:P(x) = x^{9} – 5x^{5} + 4x^{4} + 2x^{2} + 1

(i) The number of sign changes in P(x) is 2. The number of positive roots is atmost 2.

(ii) P(-x) = -x^{9} + 5x^{5} + 4x^{4} + 2x^{2} + 1. The number of sign changes in P(-x) is 1. The number of negative roots of P (x) is atmost 1. Since the difference of number of sign changes in P(-x) and number of negative zeros is even.

P(x) has one negative root.

(iii) 0 is not the zero of the polynomial P(x). So the number of real roots is almost 3.

∴ The number of imaginary roots at least 6.

Question 4.

Determine the number of positive and negative roots of the equation x^{9} – 5x^{8} – 14x^{7} = 0.

Solution:

P(x) = x^{9} – 5x^{8} – 14x^{7}. It has only one sign change.

∴ It has one positive roots.

P(-x) = -x^{9} – 5x^{8} + 14x^{7}. It has only one sign change.

It has one negative root.

∴ It has one positive and one negative roots.

Question 5.

Find the exact number of real roots and imaginary of the equation x^{9} + 9x^{7} + 7x^{5} + 5x³ + 3x.

Solution:

P(x) = x^{9} + 9x^{7} + 7x^{5} + 5x^{3} + 3x.

There is no change in the sign of P(x) and P(-x), P(x) has no positive and no negative real roots, but 0 is the root of the polynomial equation P(x).