How do you solve $x^{2} - 5 x = 0$ $x^2 - 5x =0$ ?

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Answer 1 · 2018-05-14T18:33:39

Factorize $x$ $x$ as common
$x (x - 5) = 0$ $x(x-5)=0$

Divide both sides by $x$ $x$
$x - 5 = 0$ $x-5=0$ ( as $\frac{0}{x} = 0$ $0/x=0$ )

Transfer the $5$ $5$
$x = 5$ $x=5$
Voila!

Answer 2 · 2018-05-14T18:34:03

see below

We have, $x^{2} - 5 x = 0$ $x^2-5x=0$
$x (x - 5) = 0$ $x(x-5)=0$
Thus,either $x$ $x$ can be $0$ $0$ or it can be $5$ $5$

How do you solve x2−5x=0x^2 - 5x =0?