Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.6 Text Book Back Questions and Answers, Notes.

## 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

Answer:

Given f(x) = 4x^{2} – 25

To find the zeors of f(x), put f(x) = 0

∴ 4x^{2} – 25 = 0

⇒ 4x^{2} = 25

⇒ x^{2} = \(\frac{25}{4}\)

⇒ x = ±\(\sqrt{\frac{25}{4}}\) = ±\(\frac{5}{2}\)

Hence the zeros of f(x) are \(-\frac{5}{2}, \frac{5}{2}\)

Question 2.

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

Answer:

Let f(x) = x^{3} – x^{2} – 17x – 22 = 0 —– (1)

Given that x = – 2 is a root of f(x).

∴ x + 2 is a factor of f (x)

Using synthetic division

Comparing equation (1) with the equation ax^{2} + bx + c = 0 we have

a = 1, b = – 3 , c = – 11

Question 3.

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

Answer:

x^{4} = 16

⇒ x^{4} – 16 = 0

(i.e.,) x^{4} – 4^{2} = 0

⇒ (x^{2} + 4)(x^{2} – 4) = 0

x^{2} + 4 = 0 will have no real roots

so solving x^{2} – 4 = 0

x^{2} = 4

Question 4.

Solve (2x + 1)^{2} – (3x + 2)^{2} = 0

Answer:

The given equation is (2x + 1)^{2} (3x + 2)^{2} = 0

(2x + 1 + 3x + 2) [(2x + 1) – (3x + 2)] = 0

[a^{2} – b^{2} = (a + b) (a – b)]

(5x + 3) (2x + 1 – 3x – 2) = 0

(5x + 3)(- x – 1) = 0

– (5x + 3)(x + 1) = 0

5x + 3 = 0 or x + 1 = 0

x = – \(\frac{3}{5}\) or x = – 1

∴ Solution set is { – 1, \(\frac{3}{5}\)}