Find the roots of the following quadratic equations (if they exist) by the method of completing the square. x²-4√2x+6=0
Answers
SOLUTION IS IN THE ATTACHMENT
METHOD OF COMPLETING THE SQUARE :
Step 1 - Write the given equation in standard form, ax² + bx + c = 0, a ≠ 0.
Step 2 - If the coefficient of x² is 1, go to step 3. If not, divide both sides of the equation by the coefficient of x²
Step 3 - Shift the constant term (c/a) on RHS.
Step 4- Find half the coefficient of x and square it. Add this number to both sides of the equation.
Step 5 - Write LHS in the form a perfect square and simplify the RHS.
Step 6 - Take the square root on both sides.
step 7 : Find the values of x by shifting the constant term(b/2a) on RHS from LHS.
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Solution :
i ) Compare x²-4√2x+6=0
with ax²+bx+c=0, we get
a = 1 , b = -4√2 , c = 6
Discreminant (D)
= b² - 4ac
= (-4√2)² - 4×1×6
= 32 - 24
= 8
D > 0
Therefore ,
Roots are real and distinct.
ii ) Finding roots by
Completing square method :
x² -4√2x + 6 = 0
=> x² - 2•x•2√2 = -6
=> x²-2•x•2√2+(2√2)²=-6+(2√2)²
=> ( x - 2√2 )² = -6 + 8
=> x - 2√2 = ± √2
=> x = 2√2 ± √2
Therefore ,
x = 2√2 + √2 = 3√2
Or
x = 2√2 - √2 = √2
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