# How do you solve using the completing the square method x ^2 + 13 x + 42 = 0?

Mar 20, 2016

For a very detailed example of method see
http://socratic.org/s/asVauX76

$\text{Vertex} \to \left(x , y\right) = \left(- \frac{13}{2} - \frac{1}{4}\right) = \left(- 6 \frac{1}{2} , - \frac{1}{4}\right)$
I will let you solve the other bits!

#### Explanation:

Given:$\text{ } {x}^{2} + 13 x + 42 = 0$

The process applied introduces an error that is removed by adding the correction factor of $k$

Write as:$\text{ } \left({x}^{2} + 13 x\right) + 42 = 0$

Add the correction factor of $k$

$\text{ } \left({x}^{2} + 13 x\right) + 42 + k = 0$

Take the power to outside the brackets

$\text{ } {\left(x + 13 x\right)}^{2} + 42 + k = 0$

Halve the coefficient of $13 x$

$\text{ } {\left(x + \frac{13}{2} x\right)}^{2} + 42 + k = 0$

Remove the $x$ from $\frac{13}{2} x$

$\text{ } {\left(x + \frac{13}{2}\right)}^{2} + 42 + k = 0$

The error comes from ${\left(+ \frac{13}{2}\right)}^{2} = + \frac{169}{4}$

So $k = - \frac{169}{4}$ giving:

$\text{ } {\left(x + \frac{13}{2}\right)}^{2} + 42 - \frac{169}{4} = 0$

$\text{ } {\left(x + \frac{13}{2}\right)}^{2} - \frac{1}{4} = 0$
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$\text{Vertex} \to \left(x , y\right) = \left(- \frac{13}{2} - \frac{1}{4}\right) = \left(- 6 \frac{1}{2} , - \frac{1}{4}\right)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Find x-intercepts by substituting 0 for y and solve in the normal way
y-intercept is the constant in the original equation = +42

I will let you solve for the x-intercepts 