Find the roots of the quadratic equation if they exist by Completely the square x² + 2x - 6 = 0
Answers
x² + 2x - 6 = 0
=> x² + 2x = 6
=> x² + 2.(1).x = 6
=> x² + 2.(1).x + (1)² = 6 + (1)²
here x² + 2(1)x + (1)² = (x + 1)² because it seems like a² + 2ab + b² = (a + b)²
=> (x + 1)² = 6 + 1 = 7
taking square root both sides,
=> (x + 1) = ±√7
=> x = - 1 ± √7
hence, roots are (-1 - √7) and (-1 + √7)
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 + 2x - 6 = 0
=> x² + 2x = 6
=> x² + 2x + (1)² = 6 + (1)²
=> x² + 2x +1 = 6 +1
=> x² + 2×(1)×x + (1)² = 7
=> (x + 1)² = 6 + 1 = 7
[a² + 2ab + b² = (a + b)² ]
=> (x + 1)² = 7
=> (x + 1) = ±√7
=> x = - 1 ± √7
Hence, roots of the quadratic equation are (-1 - √7) and (-1 + √7)
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