# How do you solve x^2-5x-6=0 by completing the square?

Sep 7, 2016

$x = - 1 \text{ and } x = 6$

Please do not change this solution. It is a reference solution.

#### Explanation:

$\textcolor{b l u e}{\text{End objective}}$

Given the standard form $y = a {x}^{2} + b x + c$

We end up with form $y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c + k$

The problem is that changing the original equation to this form introduces an error. So $k$ is a necessary correction. The value of which is determined by:

$a {\left(\frac{b}{2 a}\right)}^{2} + k = 0$

$\textcolor{m a \ge n t a}{\text{For this question } a = 1}$
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$\textcolor{b l u e}{\text{Completing the square}}$

Given:$\text{ } y = 0 = {x}^{2} - 5 x - 6$.......................Equation(1)

$\textcolor{b r o w n}{\text{Step 1}}$

Write as: $\left({x}^{2} - 5 x\right) - 6 + k \leftarrow \text{ At this stage "k" has no value}$

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$\textcolor{b r o w n}{\text{Step 2}}$
Move the power from ${x}^{2}$ to outside the brackets

${\left(x - 5 x\right)}^{\textcolor{m a \ge n t a}{2}} - 6 + k$

$\textcolor{p u r p \le}{\text{ Now "k" starts to take on values that change at each step}}$

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$\textcolor{b r o w n}{\text{Step 3}}$
Discard the $x$ from $5 x$ inside the brackets

${\left(x - 5\right)}^{2} - 6 + k$

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$\textcolor{b r o w n}{\text{Step 4}}$

Halve the 5 inside the brackets

${\left(x - \frac{5}{2}\right)}^{2} - 6 + k$

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$\textcolor{b r o w n}{\text{Step 5}}$

Determine the value of $k$
${\left(x \textcolor{m a \ge n t a}{- \frac{5}{2}}\right)}^{\textcolor{m a \ge n t a}{2}} - 6 + \textcolor{m a \ge n t a}{k}$......................Equation(2)

Let ${\left(\textcolor{m a \ge n t a}{- \frac{5}{2}}\right)}^{\textcolor{m a \ge n t a}{2}} + \textcolor{m a \ge n t a}{k} = 0 \text{ "rarr" Note that this is the bit:}$
color(white)("222222222222222222222222") "a(color(magenta)(b/(2a)))^(color(magenta)(2))+color(magenta)(k)=0

$\textcolor{w h i t e}{\text{ddddddddddddddddddddd}}$where in this case $a = 1$

$\implies \frac{25}{4} + k = 0 \text{ " =>" } k = - \frac{25}{4}$

So by substitution equation(2) becomes:

$y = 0 = {\left(x - \frac{5}{2}\right)}^{2} - 6 - \frac{25}{4}$

$\textcolor{b l u e}{y = 0 = {\left(x - \frac{5}{2}\right)}^{2} - \frac{49}{4}} \textcolor{w h i t e}{.} \ldots \ldots E q u a t i o n \left({2}_{a}\right)$
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$\textcolor{b r o w n}{\text{Step 6 - Determine the vertex}}$

Using $E q n \left({2}_{a}\right)$
Vertex$\to \left(x , y\right) = \left[\left(- 1\right) \times \left(- \frac{5}{2}\right) \textcolor{w h i t e}{\text{ddd"),color(white)("d}} - \frac{49}{4}\right]$
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$\textcolor{b l u e}{\text{Solving for } y = 0}$

Add $\frac{49}{4}$ to both sides

${\left(x - \frac{5}{2}\right)}^{2} = \frac{49}{4}$

Square root both sides

$x - \frac{5}{2} = \pm \sqrt{\frac{49}{4}} = \pm \frac{7}{2}$

Add $\frac{5}{2}$ to both sides

$x = \frac{5}{2} \pm \frac{7}{2}$

$x = - 1 \text{ and } x = 6$

Dec 5, 2017

As Tony B. states the answer is $x = 6$ or $x = - 1$. The explanation here uses a different method to complete the square.

#### Explanation:

In this method a different approach is used to complete the square. The idea is to show more clearly how the coefficient $5$ gets "halved".

${x}^{2} - 5 x = x \left(x - 5\right)$

These expression is the same as the first two terms of the quadratic expression originally given. We render this as a difference of squares:

$x \left(x - 5\right) = {u}^{2} - {v}^{2} = \left(u + v\right) \left(u - v\right)$

And then we match the two sets of factors:

$x = u + v$

$x - 5 = u - v$

Now add these two equations together and note that in eliminating $v$ we make a coefficient of $2$ on $u$. This is how we end up "halving the $5$:

$2 x - 5 = 2 u$ so $u = \left(x - \frac{5}{2}\right)$

Then take the difference of the original two equations to get $v$ which is just a constant, since $u$ and $x$ cancel at the same time:

$5 = 2 v$ so $v = \frac{5}{2}$; again half of $5$ appears.

This works because the original quadratic expression had a coefficient of $1$ on ${x}^{2}$. If the original quadratic expression has a different coefficient, like $2$ in $2 {x}^{2} - 10 x - 12 = 0$, we would have divided by that coefficient on ${x}^{2}$ at the beginnjng to get (in this case) ${x}^{2} - 5 x - 6 = 0$ with a coefficient of $1$ on ${x}^{2}$, and then we could proceed as above.

Then we have:

${x}^{2} - 5 x = {u}^{2} - {v}^{2} = {\left(x + \frac{5}{2}\right)}^{2} - \left(\frac{25}{4}\right)$

And the original equation becomes:

${x}^{2} - 5 x - 6 = 0$

${\left(x - \frac{5}{2}\right)}^{2} - \left(\frac{25}{4}\right) - 6 = 0$

${\left(x - \frac{5}{2}\right)}^{2} = \left(\frac{49}{4}\right)$

So we take both possible square roots noting that $\frac{49}{4}$ happens to be a perfect square:

$\left(x - \frac{5}{2}\right) = + \left(\frac{7}{2}\right)$ so $x = 6$ or

$\left(x - \frac{5}{2}\right) = - \left(\frac{7}{2}\right)$ so $x = - 1$

If we did not have a perfect square the final answer would contain square roots.