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

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

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Answer 1 · 2016-03-27T12:33:51

$x = 0$ $x=0$ or $x = \frac{1}{4}$ $x=1/4$

Explanation:

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

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

$4 x^{2} - x = {(2 x)}^{2} + 2 \times 2 x \times (- \frac{1}{4})$ $4x^2-x=(2x)^2+2xx2x xx(-1/4)$

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

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

${(2 x)}^{2} + 2 \times 2 x \times (- \frac{1}{4}) + {(- \frac{1}{4})}^{2} - {(- \frac{1}{4})}^{2} = 0$ $(2x)^2+2xx2x xx(-1/4)+(-1/4)^2-(-1/4)^2=0$ or

${(2 x - \frac{1}{4})}^{2} - {(- \frac{1}{4})}^{2} = 0$ $(2x-1/4)^2-(-1/4)^2=0$ or

$(2 x - \frac{1}{4} - \frac{1}{4}) (2 x - \frac{1}{4} + \frac{1}{4}) = 0$ $(2x-1/4-1/4)(2x-1/4+1/4)=0$ or

$(2 x - \frac{1}{2}) \cdot (2 x) = 0$ $(2x-1/2)*(2x)=0$ or

$2 (x - \frac{1}{4}) \cdot 2 x = 0$ $2(x-1/4)*2x=0$ and dividing by $4$ $4$ and rearranging we have

$x (x - \frac{1}{4}) = 0$ $x(x-1/4)=0$ i.e. $x = 0$ $x=0$ or $x = \frac{1}{4}$ $x=1/4$

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

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