# How do you solve using the completing the square method 2x^2 + 8x - 25 = 0?

Aug 24, 2016

$\implies x \approx - 6.06 \text{ and } 2.06$ to 2 decimal places

#### Explanation:

Given:$\text{ } 2 {x}^{2} + 8 x - 25 = 0$ ....................Equation(1)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Completing the square}}$

write as $y = 2 \left({x}^{2} + 4 x\right) - 25 = 0$

Introduce the correction k to compensate for the changes we are going to make. They introduce an error.

$2 \left({x}^{2} + 4 x\right) - 25 + k = 0 \leftarrow \text{ at this stage } k = 0$

Take the power of 2 outside the brackets

$2 {\left(x + 4 x\right)}^{2} - 25 + k = 0$

Remove the $x$ from $4 x$

$2 {\left(x + 4\right)}^{2} - 25 + k = 0$

Halve the 4

$2 {\left(x + 2\right)}^{2} - 25 + k = 0$............................. Equation(2)

From the part: $\textcolor{red}{2} {\left(x \textcolor{b l u e}{+ 2}\right)}^{2}$ you get the constant $\textcolor{red}{2} {\textcolor{b l u e}{\times 2}}^{2} = 8$ which is the error.

so $8 + k = 0 \implies k = - 8$ Substituting into equation(2) gives

$2 {\left(x + 2\right)}^{2} - 25 - 8 = 0. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left({2}_{a}\right)$

$2 {\left(x + 2\right)}^{2} - 33 = 0$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Determine "x_("intercepts"))

Write as${\left(x + 2\right)}^{2} = \frac{33}{2}$

Square root both sides

$x + 2 = \sqrt{\frac{33}{2}}$

$\implies x = - 2 \pm \sqrt{\frac{33}{2}}$

$\implies x \approx - 6.06 \text{ and } 2.06$ to 2 decimal places