Find the roots of the following quadratic equations (if they exist) by the method of completing the square. x²-4ax+4a²-b²=0
Answers
SOLUTION :
Given : x² - 4ax + 4a² - b² = 0
Shift the constant term on RHS
x² - 4ax = - ( 4a² - b²)
x² - 4ax = b² - 4a²
Add square of the ½ of the coefficient of x on both sides
On adding (½ of (4a))² =(2a)² both sides
x² - 4ax + (2a)² = b² - 4a² + (2a)²
Write the LHS in the form of perfect square
(x - 2a)² = b² - 4a² + 4a²
[a² - 2ab + b² = (a - b)² ]
(x - 2a)² = b²
On taking square root on both sides
(x - 2a) = √b²
(x - 2a) = ± b
On shifting constant term 2a to RHS
x = ± b + 2a
[Taking +ve sign]
x = 2a + b
[Taking - ve sign]
x = 2a - b
Hence, the roots of the given equation are 2a + b & 2a - b .
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Solution :
Compare given Quadratic
equation x²-4ax+4a²-b²=0
with Ax²+Bx+C = 0 , we get
A = 1 , B = -4a , C = 4a²-b²
i ) Discreminant (D)
= B² - 4AC
= (-4a)² - 4•1•(4a²-b²)
= 16a² - 16a² + 4b²
= 4b²
= ( 2b )²
D > o
Therefore ,
Roots are real and distinct.
ii ) Finding the roots by
Completing square method:
x² - 4ax + 4a² - b² = 0
=> x² - 4ax + 4a² = b²
=> x² -2•x•2a + (2a)² = b²
=> ( x - 2a )² = b²
=> x - 2a = ± √b²
=> x = 2a ± b
Therefore ,
x = 2a + b Or
x = 2a - b
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