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## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.6

Question 1.

Find the zeros of the polynomial function f(x) = 4x^{2} – 25

Solution:

Question 2.

If x = -2 is one root of x^{3} – x^{2} – 17x = 22, then find the other roots of equation.

Solution:

x = – 2 is one root

So applying synthetic division

Question 3.

Find the real roots of x^{4} = 16

Solution:

The given equation is x^{4} = 16

⇒ x^{4} – 16 = 0

⇒ (x^{2})^{2} – 4^{2} = 0

⇒ (x^{2} – 4) (x^{2} + 4) = 0

(x^{2} – 2^{2}) (x^{2} + 4) = 0

(x + 2)(x – 2)(x^{2} + 4) = 0

x + 2 = 0 or x – 2 = 0 or x^{2} + 4 = 0

x = -2 or x = 2

x^{2} + 4 = 0

⇒ x^{2} = – 4

⇒ x = ± √4

which is imaginaiy. Therefore, the real roots of the given equation are -2, 2.

Question 4.

Solve (2x + 1)^{2} – (3x + 2)^{2} = 0

Solution:

### Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.6 Additional Questions

Question 1.

Find the zeros of the polynomial function f(x) = 9x^{2} – 36

Solution:

Question 2.

If x = 2 is one root of x^{3} + 2x^{2} – 5x – 6 = 0 then find the other roots of the equation

Solution:

x = 2 is a root

so applying synthetic division

∴ The other factor is x^{2} + 4x + 3

Now x^{3} + 2x^{2} – 5x – 6 = (x – 2)(x^{2} + 4x + 3)

∴ x^{3} + 2x^{2} – 5x – 6 = 0 ⇒ (x – 2)(x^{2} + 4x + 3)

x – 2 = 0 or x^{2} + 4x + 3 = 0

x = 2 or (x + 1)(x + 3) = 0

⇒ x = -1 or -3

so the roots are x = -1, 2, -3