Find the roots of the following quadratic equations (if they exist) by the method of completing the square. x²-(√2+1)x+√2=0
Answers
SOLUTION :
Given : x² - (√2 + 1)x + √2 = 0
Shift the constant term on RHS
x² - (√2 + 1)x = - √2
Add square of the ½ of the coefficient of x on both sides
On adding (½ of (√2 + 1))² =(√2 + 1)/2)² both sides
x² - (√2 + 1)x + (√2 + 1)/2)² = - √2 + (√2 + 1)/2)²
Write the LHS in the form of perfect square
(x - (√2 + 1)/2)² = - √2 + [(√2² + 1² + 2√2)/4]
[a² - 2ab + b² = (a - b)² & (a + b)² = a² + 2ab + b² ]
(x - (√2 + 1)/2)² = - √2 + [ (2 + 1 + 2√2)/4]
(x - (√2 + 1)/2)² = - √2 + [ 3 + 2√2)/4]
(x - (√2 + 1)/2)² = [ (- 4√2 + 3 + 2√2)/4]
(x - (√2 + 1)/2)² = ( 3 - 2√2)/4
On taking square root on both sides
(x - (√2 + 1)/2) = √[(3 - 2√2)/4]
(x - (√2 + 1)/2) = √[√2 - 1)²]/2
[√2 - 1)² = √2² + 1² - 2√2 = 2 + 1- 2√2 = 3 - 2√2]
(x - (√2 + 1)/2) = ± [√2 - 1)]/2
On shifting constant term (√2 + 1)/2 to RHS
x = ± [√2 - 1)]/2 + (√2 + 1)/2
x = [(√2 - 1)]/2 + (√2 + 1)/2
[Taking + ve sign]
x = (√2 - 1 + √2 +1)/2
x = 2√2/2 = √2
x = √2
x = - [(√2 - 1)]/2 + (√2 + 1)/2
[Taking - ve sign]
x = (- √2 + 1 + √2 +1)/2
x = 2/2 = 1
x = 1
Hence, the roots of the given equation are √2 & 1.
HOPE THIS ANSWER WILL HELP YOU...
i )Compare x²-(√2-1)x+√2=0
with ax²+bx+c=0 , we get
a = 1 , b = -(√2-1), c = √2
Discreminant (D)
= b² - 4ac
= [-(√2+1)]² - 4×1×√2
= ( 1 + √2 )² - 4×1×√2
= ( 1 - √2 )²
D > 0
Therefore ,
Roots are real and distinct.
ii ) Finding roots by
Completing square method:
x²-(√2+1)x+√2=0
=> x² - 2•x•[(√2+1)/2]= -√2
=> x² - 2•x•[(√2+1)/2]+[(√2+1)/2]²
= -√2 + [( √2 + 1 )/2]²
=> [ x - (√2+1)/2]² =-√2+(√2+1)²/4
=> [x-(√2+1)/2]² = [-4√2+(√2+1)²]/4
=> [x-(√2+1)/2]² = (√2-1)²/4
=> [x-(√2+1)/2] = ± (√[(√2-1)/2]²
=> x = (√2+1)/2 ± (√2-1)/2
=> x= [(√2+1)±(√2-1)]/2
Now ,
x = [√2+1+√2-1]/2
= 2√2/2
= √2
Or
x = [√2+1-√2+1]/2
= 2/2
= 1
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