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 = - 31/16
which is not possible at the square of a real number cannot be negative.
Therefore, roots of the equation does not exist.
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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
= -31
D < 0
Therefore ,
Roots are imaginary.
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 = ± √(-31/16)
=> x = -1/4 ± √(-31)/4
=> x = [-1 ± √(-31) ]/4
Therefore ,
x = [-1+√(-31)]/4
Or
x = [-1-√(-31)]/4
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