How do you solve 2x^2 + 32x + 12 = 02x2+32x+12=0 using completing the square?

1 Answer
Gió
Jun 24, 2015

I found:
x_1=-8+sqrt(58)x1=−8+√58
x_2=-8-sqrt(58)x2=−8−√58

Explanation:

Write it as (dividing by 22):
x^2/2+32/2x+12/2=0x22+322x+122=0
x^2+16x+6=0x2+16x+6=0
x^2+16x=-6x2+16x=−6
Add and subtract 6464:
x^2+16x+64-64=-6x2+16x+64−64=−6
x^2+16x+64=-6+64x2+16x+64=−6+64
(x+8)^2=58(x+8)2=58
x+8=+-sqrt(58)x+8=±√58
x=-8+-sqrt(58)x=−8±√58
so:
x_1=-8+sqrt(58)x1=−8+√58
x_2=-8-sqrt(58)x2=−8−√58

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