1. Determine the nature of the roots of the equation.
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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
I HOPE ITS HELP YOU DEAR,
THANKS
(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
I HOPE ITS HELP YOU DEAR,
THANKS
Answered by
0
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.
••••
=> 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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