Find the roots of the following quadratic equations (if they exist) by the method of completing the square. 2x² + x − 4 = 0
Answers
SOLUTION :
Given : 2x² + x - 4 = 0
On dividing the whole equation by 2,
(x² + x/2 - 4/2) = 0
(x² + x/2 - 2) = 0
Shift the constant term on RHS
x² + x/2 = 2
Add square of the ½ of the coefficient of x on both sides
On adding (½ of 1/2)² = (1/4)² both sides
x² + x/2 + (1/4)²= 2 + (1/4)²
Write the LHS in the form of perfect square
(x + 1/4)² = 2 + 1/16
[a² + 2ab + b² = (a + b)²]
(x + 1/4)² = (2 × 16 + 1)/16
(x + 1/4)² = (32 + 1)/16
(x + 1/4)² = 33/16
On taking square root on both sides
(x + ¼) = √(33/16)
(x + ¼) = ± √(33)/4
On shifting constant term (1/4) to RHS
x =± √(33)/4 - 1/4
x = √(33)/4 - 1/4
[Taking +ve sign]
x = (√33 - 1)/4
x =± √(33)/4 - 1/4
x = - √(33)/4 - 1/4
[Taking - ve sign]
x = (- √33 - 1)/4
Hence, the roots of the given equation are (√33 - 1)/4 & (- √33 - 1)/4
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Solution :
Compare given Quadratic
equation 2x²+x-4 = 0 by
ax² + bx + c = 0, we get
a = 2 , b = 1 , c = -4 ,
i ) Discreminant ( D )
= b² - 4ac
= 1² - 4×2×(-4)
= 1 + 32
= 33
D > 0
Therefore ,
Roots are real and distinct .
ii ) Finding roots by
completing square method:
2x² + x - 4 = 0
Divide each term by 2 , we
get
x² + x/2 - 2 = 0
=> x² + 2•x•(1/4) = 2
=> x²+2•x•(1/4)+(1/4)²=2+(1/4)²
=> ( x + 1/4 )² = 2 + 1/16
=> ( x + 1/4 )² = ( 32 + 1 )/16
=> ( x + 1/4 )² = 33/16
=> x + 1/4 = ± √(33/16)
=> x = -1/4 ± √(33)/4
=> x = ( -1 ± √33 )/4
Therefore ,
x = ( -1 + √33 )/4
Or
x = ( -1 - √33 )/4
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