How do you solve $9x^2 = 27x$ ?

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Answer 1 · 2016-07-16T04:35:21

$x = 3$ and $x = 0$

We can rewrite $9x^2 = 27x$ by subtracting $27x$ from both sides and make it equal to $0$ .

$9x^2 = 27x$

$9x^2 - 27x = 0$

Factoring out a $9$ and $x$ we now get

$9x(x-3) = 0$

We now have two separate factors, namely $9x = 0$ and $x-3 = 0$ , so we can simply solve for $x$ .

$9x = 0 -> x = 0$

$x-3=0 -> x =3$

So our solutions are $x = 3$ and $x = 0$ .

To check our answer, we can also graph $9x^2-27x$ .

graph{9x^2-27x [-2.82, 4.973, -1.955, 1.942]}

Answer 2 · 2016-07-16T04:37:34

$x=0$ or $x=3$

$9x^2=27x$ $hArr$

$9x^2-27x=0$ or

$9x(x-3)=0$ or dividing by $9$

$x(x-3)=0$ i.e. either $x=0$ or $x-3=0$

Hence $x=0$ or $x=3$

How do you solve 9x^2 = 27x9x^2 = 27x?