How do you find the roots of $x^2+x=56$ ?

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Answer 1 · 2016-05-14T19:50:35

$x = 7$ or $x = -8$

Note that $x^2+x = x(x+1)$ and $56 = 7 * 8$

So $x = 7$ is a root.

This is a quadratic, so it has another root.

$56 = (-8)*(-7)$

So $x = -8$ is the other root.

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Alternative method

Alternatively, we can subtract $56$ from both sides to get:

$x^2+x-56 = 0$

This is then in the form $ax^2+bx+c = 0$ with $a=1$ , $b=1$ and $c = -56$ .

This has roots given by the quadratic formula:

$x = (-b+-sqrt(b^2-4ac))/(2a)$

$=(-1+-sqrt(1^2-(4*1*(-56))))/(2*1)$

$=(-1+-sqrt(1+224))/2$

$=(-1+-sqrt(225))/2$

$=(-1+-15)/2$

We find:

$(-1+15)/2 = 14/2 = 7$

$(-1-15)/2 = (-16)/2 = -8$

How do you find the roots of x^2+x=56x^2+x=56?