How do you solve the equation $3x^2-x+5=x^2+3x-14$ by completing the square?

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Answer 1 · 2017-02-15T17:00:12

$2(x-1)^2+17 = 0$

Combine like terms by moving all terms on the right to the left:
$(3x^2-x^2) +(-x-3x) +(5+14) = 0$

Simplify: $2x^2 -4x +19 = 0$

Complete the square:

Combine the x-terms & factoring: $2(x^2-2x) +19 = 0$
Half the x-term $-2x: 1/2(-2) = -1$ , Used in $(x-1)^2$
Square the halved x-term from step 2: $(-1)^2 = 1$ , and multiply by $2$ , the factored number in front of the square.
Complete the square & subtract the added term: $2(x-1)^2 +19 - 2(1) = 0$
Simplify: $2(x-1)^2 +17 = 0$

How do you solve the equation 3x^2-x+5=x^2+3x-143x^2-x+5=x^2+3x-14 by completing the square?