# How do you solve x^2+x=20?

Jun 24, 2017

$x = 4 \text{ }$ or $\text{ } x = - 5$

#### Explanation:

Here are a few methods:

$\textcolor{w h i t e}{}$
Method 1 - Fishing for factors

Note that ${x}^{2} + x = x \left(x + 1\right)$, so the given equation can be rewritten:

$x \left(x + 1\right) = 20$

So we want to find a number $x$, which when multipled by $x + 1$ gives $20$.

It should not take too much guessing to find:

$\textcolor{b l u e}{4} \left(\textcolor{b l u e}{4} + 1\right) = \left(4\right) \left(5\right) = 20$

So $x = 4$ is a solution.

This is a quadratic equation, so should normally have two solutions, so what is the other?

With a bit of thought, we find:

((color(blue)(-5))((color(blue)(-5))+1) = (-5)(-4) = 20

So $x = - 5$ is the other solution.

$\textcolor{w h i t e}{}$
Method 2 - Completing the square

Given:

${x}^{2} + x = 20$

Note that in general:

${\left(x + p\right)}^{2} = {x}^{2} + 2 p x + {p}^{2}$

So to make ${x}^{2} + x$ into a perfect square, we need to choose $p = \frac{1}{2}$ and add ${p}^{2} = \frac{1}{4}$ to both sides of our equation to get:

${x}^{2} + x + \frac{1}{4} = \frac{81}{4}$

Both sides are now perfect squares:

${\left(x + \frac{1}{2}\right)}^{2} = {\left(\frac{9}{2}\right)}^{2}$

Hence:

$x + \frac{1}{2} = \pm \frac{9}{2}$

Subtracting $\frac{1}{2}$ from both sides:

$x = - \frac{1}{2} \pm \frac{9}{2}$

That is:

$x = 4 \text{ }$ or $\text{ } x = - 5$

$\textcolor{w h i t e}{}$

Given:

${x}^{2} + x = 20$

Subtract $20$ from both sides to get:

${x}^{2} + x - 20 = 0$

This is in the standard form:

$a {x}^{2} + b x + c = 0$

with $a = 1$, $b = 1$ and $c = - 20$.

It has solutions given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- 1 \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(1\right) \left(- 20\right)}}{2 \cdot 1}$

$\textcolor{w h i t e}{x} = \frac{- 1 \pm \sqrt{1 + 80}}{2}$

$\textcolor{w h i t e}{x} = \frac{- 1 \pm 9}{2}$

That is:

$x = 4 \text{ }$ or $\text{ } x = - 5$

Jul 21, 2017

$4 , - 5$

#### Explanation:

Method 4. Solving by new Transforming Method (Google Search)
$y = {x}^{2} + x - 20 = 0$
Find 2 real roots, that have opposite signs (ac < 0), knowing the sum (- b = - 1) and the product (c = - 20).
They are: $4 \mathmr{and} - 5$