Find the roots of the following quadratic equations (if they exist) by the method of completing the square. √2x²-3x-2√2=0
Answers
SOLUTION :
Given : √2x² - 3x - 2√2 = 0
On dividing the whole equation by √2,
(x² - 3x/√2 - 2) = 0
Shift the constant term on RHS
x² - 3x/√2 = 2
Add square of the ½ of the coefficient of x on both sides
On adding (½ of 3/√2)² = (3/2√2)² both sides
x² - 3x/√2 + (3/2√2)² = 2 + (3/2√2)²
Write the LHS in the form of perfect square
(x - 3/2√2)² = 2 + 9/8
[a² - 2ab + b² = (a - b)²]
(x - 3/2√2)² = (2 × 8 + 9)/8
(x - 3/2√2)² = (16 + 9)/8
(x - 3/2√2)² = 25/8
On taking square root on both sides
(x - 3/2√2) = √(25/8)
(x - 3/2√2) = ± 5/√8
(x - 3/2√2) = ± 5/2√2
On shifting constant term (- 3/2√2) to RHS
x = ± 5/2√2 + 3/2√2
x = (5 + 3)/2√2
[Taking + ve sign]
x = 8/2√2
x = 4/√2
On rationalising the denominator
x = (4 × √2) / √2 × √2
x = 4√2 /2
x = 2√2
x = (- 5 + 3)/2√2
[Taking - ve sign]
x = - 2 /2√2
x = - 1/√2
Hence, the roots of the given equation are 2√2 & - 1/√2 .
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Solution :
Compare given Quadratic
equation √2x²-3x-2√2=0
with ax²+bx+c=0 , we get
a = √2 , b = -3 , c = -2√2
i ) Discreminant (D)
= b² - 4ac
= (-3)² - 4×(√2)(-2√2)
= 9 + 16
= 25
= 5²
D > 0
Therefore ,
Roots are real and distinct.
ii ) Finding roots by
completing square method:
√2x² - 3x - 2√2 = 0
Divide each term by √2,
we get
=> x² - (3/√2)x - 2 = 0
=> x² - 2•x•(3/2√2) = 2
=> x²-2•x•(3/2√2)+(3/2√2)²=2+(3/2√2)²
=> ( x - 3/2√2 )² = 2 + 9/8
=> ( x - 3/2√2 )² = (16 + 9 )/8
=> ( x - 3/2√2 )² = 25/8
=> x - 3/2√2 = ± √(25/8)
=> x = 3/2√2 ± 5/2√2
=> x = ( 3 ± 5 )/2√2
Therefore ,
x = (3+5)/2√2 Or x=(3-5)/2√2
=> x = 8/2√2 Or x = -2/2√2
=> x = 4/√2 Or x = -1/√2
=> x=(2•√2•√2)/√2 Or x = -1/√2
=> x = 2√2 Or x = -1/√2
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