# If a number is added to its square, the result is 56. How do you find the number?

Mar 29, 2018

$\textcolor{b l u e}{a = - 8}$

$\textcolor{b l u e}{a = 7}$

#### Explanation:

Let the number be $a$:

$\therefore$

${a}^{2} + a = 56$

We now solve the equation:

${a}^{2} + a - 56 = 0$

Factor:

$\left(a + 8\right) \left(a - 7\right) = 0 \implies a = - 8 \mathmr{and} a = 7$

Mar 29, 2018

Call the number $n$, then we know $n + {n}^{2} = 56$. We can write this as a quadratic equation in standard form: ${n}^{2} + n - 56 = 0$. Using the quadratic formula, or by factorising, we find $n = - 8$ or $7$..

#### Explanation:

I normally use the quadratic formula: $n = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

It's not hard to remember, and it always works.

For this particular one, though, I will factorise instead. I need two numbers that multiply to give $- 56$ and add to give $+ 1$. Two numbers that fit are $8$ and $- 7$.

So $\left(n + 8\right) \left(N - 7\right) = 0$

If you expand that you will get back our quadratic.

The two 'roots' - the solutions of the quadratic - are the values of $n$ that make each bracket go to zero.

Those roots are $- 8$ and $7$. Either value fulfils the criteria in the question.