Find the roots of the quadratic equation if they exist by Completely the square x² + 7x - 13 = 0
Answers
x² + 7x - 13 = 0
=> x² + 7x = 13
=> x² + 2.(7/2)x = 13
=> x² + 2.(7/2).x + (7/2)² = 13 + (7/2)²
=> (x + 7/2)² = 13 + 49/4
=> (x + 7/2)² = (52 +49)/4 = 101/4
=> (x + 7/2) = ±√(101)/2
=> x = -7/2 ± √(101)/2
=> x = (-7 ± √101)/2
hence, roots of quadratic equation are (-7 - √101)/2 and (-7 + √101)/2
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. SOLUTION :
Given :
x² + 7x - 13 = 0
=> x² + 7x = 13
=> x² + 7x +(7/2)² = 13 +(7/2)²
=> x² + 2×(7/2)x +(7/2)² = 13 +(7/2)²
=> (x + 7/2)² = 13 + 49/4
=> (x + 7/2)² = (52 +49)/4 = 101/4
=> (x + 7/2)² = 101/4
=> (x + 7/2) = ±√(101)/2
=> x = -7/2 ± √(101)/2
=> x = -7/2 + √101/2
=> x = -7/2 - √101 /2
Hence, roots of quadratic equation are (-7 - √101)/2 & (-7 + √101)/2
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