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

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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) = 4x² – 25
Solution:

Question 2.
If x = -2 is one root of x³ – x² – 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 x⁴ = 16
Solution:
The given equation is x⁴ = 16
⇒ x⁴ – 16 = 0
⇒ (x²)² – 4² = 0
⇒ (x² – 4) (x² + 4) = 0
(x² – 2²) (x² + 4) = 0
(x + 2)(x – 2)(x² + 4) = 0
x + 2 = 0 or x – 2 = 0 or x² + 4 = 0
x = -2 or x = 2
x² + 4 = 0
⇒ x² = – 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) = 9x² – 36
Solution:

Question 2.
If x = 2 is one root of x³ + 2x² – 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 x² + 4x + 3
Now x³ + 2x² – 5x – 6 = (x – 2)(x² + 4x + 3)
∴ x³ + 2x² – 5x – 6 = 0 ⇒ (x – 2)(x² + 4x + 3)
x – 2 = 0 or x² + 4x + 3 = 0
x = 2 or (x + 1)(x + 3) = 0
⇒ x = -1 or -3
so the roots are x = -1, 2, -3