How do you solve $x^2-x=12$ ?

Question

6310 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-02-03T13:58:15

We can use the Sum-Product method.

Bring everything to one side:
$x^2-x=12->x^2-x-12=0$

We now have a quadratic equation of the form
$ax^2+bx+c=0$ where $a=1, b=-1 and c=-12$

We find two numbers that will give $c=-12$ as a product and $b=-1$ as a sum (or difference).
We can try $1*12, 2*6,3*4$

$3and4$ will fit, with a $-$ sign to the $4$ , as that would make the sum $3-4=-1$ and the product $3*-4=-12$

Now we can rewrite the equation:
$(x+3)(x-4)=0$

Which leaves us with two possibilities:
$(x+3)=0->x=-3$ OR $(x-4)=0->x=4$

Answer:
$x=-3 or x=4$

How do you solve x^2-x=12x^2-x=12?