# How do you solve using the completing the square method x^2 + 3x - 10 = 0?

Apr 13, 2016

#### Answer:

$\textcolor{b l u e}{{y}_{\text{intercept}} = - 10}$

$\textcolor{b l u e}{\text{Vertex} \to \left(x , y\right) \to \left(- \frac{3}{2} , - \frac{49}{4}\right)}$

$\textcolor{b l u e}{{x}_{\text{intercepts}} = - 5 \mathmr{and} + 2}$

#### Explanation:

Given:$\text{ } {x}^{2} + 3 x - 10 = 0$.......................(1)

color(blue)("Determine "y_("intercept"))

Read directly from equation (1)

$\textcolor{b l u e}{{y}_{\text{intercept}} = - 10}$

$\textcolor{b r o w n}{\text{'~~~~~~~~~~~~~ Tip! ~~~~~~~~~~~~~~~~~~~~}}$
$\textcolor{g r e e n}{\text{As the equation is already in the form }}$

$\textcolor{g r e e n}{y = a \left({x}^{2} + \frac{b}{a} x\right) + c}$

$\textcolor{g r e e n}{\text{In this case } a = 1}$

$\textcolor{g r e e n}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a} = - \frac{3}{2}}$

$\textcolor{b r o w n}{\text{'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$

$\textcolor{b l u e}{\text{Step 1}}$
Write equation (1) as $\left({x}^{2} + 3 x\right) - 10 = 0$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 2}}$

Add the adjustment constant $k$
$\left({x}^{2} + 3 x\right) - 10 + k = 0$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 3}}$
Move the power from ${x}^{2}$ to outside the bracket
${\left(x + 3 x\right)}^{2} - 10 + k = 0$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 3}}$

Remove the $x \text{ from } 3 x$
${\left(x + 3\right)}^{2} - 10 + k = 0$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 4}}$

Multiply the 3 inside the bracket by $\frac{1}{2}$
${\left(x + \frac{3}{2}\right)}^{2} - 10 + k = 0$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 4}}$
If we multiply out the bracket we end up with an additional term to those in the original equation. That term is ${\left(\frac{3}{2}\right)}^{2}$ derived from (?+3/2)(?+3/2)=(3/2)^2. This term must be removed which is achieved by making $k = - {\left(\frac{3}{2}\right)}^{2}$

So now we have

$\textcolor{b r o w n}{{\left(x + \frac{3}{2}\right)}^{2} - 10 + k = 0} \textcolor{g r e e n}{\to {\left(x + \frac{3}{2}\right)}^{2} - 10 - {\left(\frac{3}{2}\right)}^{2} = 0}$

$\textcolor{b l u e}{\text{Completing the square } \to} \textcolor{m a \ge n t a}{{\left(x + \frac{3}{2}\right)}^{2} - \frac{49}{4} = 0}$......................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
" "color(blue)("Vertex" -> (x,y)->(-3/2,-49/4))

From equation (2)

${\left(x + \frac{3}{2}\right)}^{2} = \frac{49}{4}$

Square rooting both sides

$x + \frac{3}{2} = \pm \frac{\sqrt{49}}{\sqrt{4}}$

 color(blue)(x_("intercepts")=-3/2+-7/2 = -5 or +2)# 