How do you solve $2x^2+5x-1=0$ by completing the square?

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Answer 1 · 2018-04-19T12:03:33

$2(x+1.25)^2-4.125=0$

We first take the first two terms and factor out the coefficient of $x^2$ :
$(2x^2)/2+(5x)/2=2(x^2+2.5x)$

Then we divide by $x$ , half the integer and square what remains:
$2(x^2/x+2.5x/x)2=2(x+2.5)$
$2(x+2,5/2)=2(x+1.25)$
$2(x+1..25)^2$

Expand out the bracket:
$2x^2+2.5x+2.5x+2(1.25^2)=2x^2+5x+3.125$

Make it equal the original equations:
$2x^2+5x+3.125+a=2x^2+5x-1$

Rearrange to find $a$ :
$a=-1-3.125=-4.125$

Put in $a$ to the factorised equation:
$2(x+1.25)^2-4.125=0$

How do you solve 2x^2+5x-1=02x^2+5x-1=0 by completing the square?