How do you solve $4x^2 - x = 0$ by completing the square?

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Answer 1 · 2016-03-27T12:33:51

$x=0$ or $x=1/4$

As $a^2+2ab+b^2=(a+b)^2$

Trying to convert $4x^2-x$ in form $a^2+2ab$ we can have

$4x^2-x=(2x)^2+2xx2x xx(-1/4)$

This means we should add $(-1/4)^2$ on both sides for completing the square.

Hence, $4x^2-x=0$ can be written as

$(2x)^2+2xx2x xx(-1/4)+(-1/4)^2-(-1/4)^2=0$ or

$(2x-1/4)^2-(-1/4)^2=0$ or

$(2x-1/4-1/4)(2x-1/4+1/4)=0$ or

$(2x-1/2)*(2x)=0$ or

$2(x-1/4)*2x=0$ and dividing by $4$ and rearranging we have

$x(x-1/4)=0$ i.e. $x=0$ or $x=1/4$

How do you solve 4x^2 - x = 04x^2 - x = 0 by completing the square?