Now take x = 2, a = 1 and b = 3, verify the identity (x + a) (x + b)

1.	Now take x = 2, a = 1 and b = 3, verify the identity (x + a) (x + b) s x + (a + b)x + ab. i) What do you observe? Is LHS = RHS?ii) Take different values for x, a and b for verification of the above identity. iii) Is it always LHS = RHS for all values of a and b?
Answer» i) (x + a) (x + b) = x² + (a + b)x + ab x = 2, a = 1, b = 3 then ⇒ (2 + 1) (2 + 3) = 2² + x(1 + 3)² + 1 × 3 ⇒ 3 × 5 = 4 + 4x² + 3 ⇒ 15 = 4 + 8 + 3 ⇒ 15 = 15 ∴ LHS = RHS ii) x = 0, a = 1, b = 2 then ⇒ (0 + 1) (0 + 2) = 0² + (1 + 2) 0 + 1 × 2 ⇒ 1 × 2 = 0 + 0 + 2 ⇒ 2 = 2 ∴ LHS = RHS for different values of x, a, b. iii) LHS = RHS for all the values of a, b.

Now take x = 2, a = 1 and b = 3, verify the identity (x + a) (x + b) s x + (a + b)x + ab. i) What do you observe? Is LHS = RHS?ii) Take different values for x, a and b for verification of the above identity. iii) Is it always LHS = RHS for all values of a and b?

Answer»

i) (x + a) (x + b) = x² + (a + b)x + ab

x = 2, a = 1, b = 3 then

⇒ (2 + 1) (2 + 3) = 2² + x(1 + 3)² + 1 × 3

⇒ 3 × 5 = 4 + 4x² + 3

⇒ 15 = 4 + 8 + 3

⇒ 15 = 15

∴ LHS = RHS

ii) x = 0, a = 1, b = 2 then

⇒ (0 + 1) (0 + 2) = 0² + (1 + 2) 0 + 1 × 2

⇒ 1 × 2 = 0 + 0 + 2

⇒ 2 = 2

∴ LHS = RHS for different values of x, a, b.

iii) LHS = RHS for all the values of a, b.