How do you solve by completing the square: $7w² + 6w + 7 = 0$ ?

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Answer 1 · 2015-03-30T10:45:51

In this way:

$7w^2+6w+7=0rArr7(w^2+6/7w)+7=0rArr$

$7(w^2+6/7w+(3/7)^2-(3/7)^2)+7=0rArr$

$7(w^2+6/7w+(3/7)^2)-7*9/49+7=0rArr$

$7(w+3/7)^2-9/7+7=0rArr$

$7(w+3/7)^2=9/7-7rArr$

$7(w+3/7)^2=(9-63)/7rArr$

$7(w+3/7)^2=-54/7$

that is an impossible equation, because a square can't be negative!

How do you solve by completing the square: 7w² + 6w + 7 = 07w² + 6w + 7 = 0?