Question

1. Determine the nature of the roots of the equation.

Answer 1

HELLO DEAR,



(x - 2a)(x - 2b) = 4ab

x² - 2bx - 2ax + 4ab = 4ab

x² - 2(a + b)x = 0

where, a = 1 ,b = -2(a + b) , c = 0

now,
D = b² - 4ac

D = [-2(a + b)]² - 4(1)(0)

D = 4(a + b)²
here, D can't be negative because value of a and b can (-ve or +ve) so Square of (-ve)no. or (+ve)no. always gives positive no.

So, the roots are $\underline{real}$


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Given (x-2a)(x-2b)= 4ab

=> x(x-2b)-2a(x-2b)-4ab = 0

=> x²-2bx-2ax+4ab-4ab=0

=> x² - 2bx - 2ax = 0

=> x² - 2( a + b ) x = 0

Compare above equation with

Ax² + Bx + C = 0 , we get

A = 1 ,

B = -2( a + b ),

C = 0 ,

Discreminant = B² - 4AC

= [ -2( a + b ) ]² - 4 × 1 × 0

= [ 2( a + b ) ]²

≥ 0

Discreminant ≥ 0

Therefore ,

Roots are real and distinct.

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