Find the roots of the following quadratic equations (if they exist) by the method of completing the square. 4x²+4√3x+3=0
Answers
SOLUTION :
Given : 4x² + 4√3x + 3 = 0
On dividing the whole equation by 4,
(x² + √3x + 3/4) = 0
Shift the constant term on RHS
x² + √3x = - 3/4
Add square of the ½ of the coefficient of x on both sides
On adding (½ of √3)² = (√3/2)² both sides
x² + √3x + (√3/2)² = -¾ + (√3/2)²
Write the LHS in the form of perfect square
(x² + √3/2)² = -¾ + 3/4
[a² + 2ab + b² = (a + b)²]
(x² + √3/2)² = 0
On taking square root on both sides
(x² + √3/2) = ± 0
On shifting constant term (√3/2) to RHS
x = - √3/2
Hence, the roots of the given equation are - √3/2 & - √3/2 .
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Solution :
Compare given Quadratic
equation 4x²+4√3x+3=0
with ax²+bx+c=0 , we get
a = 4 , b = 4√3 , c = 3
i ) Discreminant ( D )
= b² - 4ac
= ( 4√3 )² - 4×4×3
= 48 - 48
= 0
D = 0
Therefore ,
Roots are real and equal.
ii ) Finding roots by
Completing square method:
4x² + 4√3x + 3 = 0
Divide each term by 4 ,
we get ,
x² + √3x + 3/4 = 0
=> x²+2•x•(√3/2) = -3/4
=> x²+2•x•(√3/2)+(√3/2)²
= -3/4 + ( √3/2 )²
=> ( x + √3/2 )² = -3/4 + 3/4
=> ( x + √3/2 )² = 0
=> x + √3/2 = 0 Or x+√3/2 = 0
=> x = -√3/2 Or x = -√3/2
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