# The square of a positive number is 56 more than the number itself. What is the number?

Jan 9, 2017

The number is $8$

#### Explanation:

We need to take this one phrase at a time to develop our equation.

First, the square of a positive number can be written as:

${x}^{2}$

In mathematics the word "is"means "=" so we can now write:

${x}^{2} =$

and "56 more than the number itself" finishes the equation as:

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

We can now transform this into a quadratic:

${x}^{2} - \textcolor{red}{56 - x} = 56 + x - \textcolor{red}{56 - x}$

${x}^{2} - x - 56 = 0$

We can now factor the quadratic:

$\left(x - 8\right) \left(x + 7\right) = 0$

Now we can solve each term for $0$

$x + 7 = 0$

$x + 7 - 7 = 0 - 7$

$x + 0 = - 7$

$x = - 7$ - this cannot be the answer because the question asked for a positive integer.

$x - 8 = 0$

$x - 8 + 8 = 0 + 8$

$x - 0 = 8$

$x = 8$

The number is $8$

Jan 9, 2017

$8$

#### Explanation:

Let the unknown value be $x$

This is a quadratic in disguise.

${x}^{2} = x + 56 \text{ "=>" } {x}^{2} \textcolor{red}{- x} - 56 = 0$

The $\textcolor{red}{x}$ has the coefficient of -1. This means that the whole number factors of 56 have a difference of -1.

$\sqrt{56} \approx 7.5$

Try $\left(- 8\right) \times \left(+ 7\right) = - 56 \text{ and } 7 - 8 = - 1$ so we have found the factors

${x}^{2} - x - 56 = \left(x - 8\right) \left(x + 7\right) = 0$

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The question stipulates that the number is positive so we select $x = + 8$
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$\textcolor{b l u e}{\text{Check}}$

${x}^{2} = x + 56 \text{ "->" } {8}^{2} \to 8 + 56$

$\text{ } 64 \to 64$