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## 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} + x^{3} + 7x^{2} + 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 atmost 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 zeros of the polynomials x^{2} – 5x + 6 and x^{2} – 5x + 16. Also, draw a rough sketch of the graphs.

Solution:

P(x) = x^{2} – 5x + 6

The number of sign changes in P(x) is 2.

P(x) has atmost 2 positive roots. P(-x) = x^{2} + 5x + 6.

The number of sign changes in P(-x) is 0.

∴ P (x) has no negative roots. P (x) = x^{2} – 5x + 16

Question 3.

Show that the equation x^{9} – 5x^{5} + 4x^{4} + 2x^{2} +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:

x^{9} – 5x^{8} – 14x^{7} = 0

P (x) = x^{9} – 5x^{8} – 14x^{7}. The number of sign changes is P(x) is 1.

The number of positive roots is 1. P (-x) = -x^{9} – 5x^{8} + 14x^{7}

The number of sign changes is P(-x) is one. The number of negative zero of P(-x) is 1. It is clear that 0 is a root of the equation.

∴ The number of the imaginary roots is at least 6.

Question 5.

Find the exact number of real zeros and imaginary of the polynomial x^{9} + 9x^{7} + 7x^{5} + 5x^{3} + 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).

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

Question 1.

Find the maximum possible number of real roots of the equation, x^{5} – 6.x^{2} – 4x + 5 = 0.

Solution:

Let f(x) = x^{5} – 6x^{2} – 4x + 5

Check the terms when it changes sign

Number of changes = 2

∴ Maximum number of positive real roots = 2

f(-x) = (- x)^{5} – 6 (- x)^{2} – 4 (-x) + 5

= -x^{5} – 6x^{2} + 4x + 5

Check the terms when it changes signs.

Number of changes = 1

∴ Maximum number of negative real roots = 1

∴ Total maximum number of real roots = 2 + 1 = 3

Question 2.

Find the number of real roots of the equation, x^{2} + 5 |x| + 6 = 0.

Solution:

|x^{2}| – 5|x| + 6 = 0

(|x| -2) (|x| -3) = 0

It has four real roots. The real roots are 2, -2, 3, -3

Question 3.

Find the real roots of the equation, x^{2} + 5 |x| + 6 = 0.

Solution:

x^{2} + 5 |x| + 6 = 0

Case (i) If x ≥ 0

x^{2} + 5x + 6 = 0

(x + 2) (x + 3) = 0

x = -2 and x = -3

Case (ii) If x < 0

x^{2} – 5x + 6 = 0

(x – 2) (x – 3) = 0

x = 2 and x = 3

Question 4.

Solve x^{4} – 4x^{2} + 8x + 35 = 0. Given (2 + \(i \sqrt{3}\)) is a root.

Solution:

⇒ x^{2} – 4x + 7 is a factor of P (x). Dividing the polynomial P(x) = 0 by x^{2} – 4x + 7.

We get x^{2} + 4x + 5 = 0 is a other factor. The roots of x^{2} + 4x + 5 = 0 are

Question 5.

Solve x^{4} – 5x^{3} + 4x^{2} + 8x – 8 = 0. Given (1 – \(\sqrt{5}\)) is a root of the polynomial equation.

Solution:

Since (1 – \(\sqrt{5}\)) is a root of the polynomial P(x) = 0

(1 + \(\sqrt{5}\)) is also a root of P (x) = 0

⇒ x^{2} – [(1 + \(\sqrt{5}\)) + (1 – \(\sqrt{5}\))] x + (1 + \(\sqrt{5}\) ) (1 – \(\sqrt{5}\) ) = 0 is a factor of P(x) = 0 ⇒ x^{2} – 2x – 4 = 0 is a factor of P(x) = 0.

Dividing the polynomial by x^{2} – 2x – 4 = 0

We get the other factor x^{2} – 3x + 2 = 0

The roots of x^{2} – 3x + 2 = 0

(x – 2) (x – 1) = 0

x = 1, 2

The roots are 1, 2, 1 ± \(\sqrt{5}\)

Question 6.

Find a polynomial equation of the lowest degree with rational co-efficient having \(\sqrt{3}\), (1 – 2i) as two of its roots.

Solution:

When \(\sqrt{3}\) is a root, – \(\sqrt{3}\) will also be a root.

Now the quadratic equation with \(\sqrt{3}\) , – \(\sqrt{3}\) are roots is x^{2} – (\(\sqrt{3}\) – \(\sqrt{3}\))x + (\(\sqrt{3}\))(-\(\sqrt{3}\)) = 0

(i.e) x^{2} – 3 = 0

When 1 – 2i is a root, 1 + 2i will be another root.

Now the quadratic equation with roots 1 – 2i and 1 + 2i is

x^{2} – (1 – 2i + 1 + 2i)x + (1 – 2i)(1 + 2i) = 0

(i.e) x^{2} – 2x + 5 = 0

∴ The equation with roots ± \(\sqrt{3}\) and 1 ± 2i is (x^{2} – 3) (x^{2} – 2x + 5) = 0

(i.e) x^{4} – 2x^{3} + 5x^{2} – 3x^{2} + 6x – 15 = 0 ,

(i.e) x^{4} – 2x^{3} + 2x^{2} + 6x – 15 = 0