# How do you solve 2x^2-12x=-14 by completing the square?

Apr 26, 2016

$\text{ "x~~4.414" and } x \approx 1.586$ to 3 decimal places

#### Explanation:

Write as $2 {x}^{2} - 12 x + 14 = 0$

$\textcolor{red}{\text{Method for completing the square in detail}}$

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

What process we are about to do introduces a value that is not in the original equation. So we mathematically compensate for this by the inclusion of a correction value. This correction value would turn the introduced error into 0 if we were to carry out the addition.

Suppose we had $2 z + 56$. Then suppose we added 4 which is viewed as an error. However if we write $2 z + 56 + 4 - 4$ we have corrected the error

Let $k$ be the constant of correction

Write as $\text{ } 2 \left({x}^{2} - 6 x\right) + 14 + k = 0$

Move the power of 2 to outside the bracket

$\text{ } 2 {\left(x - 6 x\right)}^{2} + 14 + k = 0$

divide the 6 from $6 x$ by 2

$\text{ } 2 {\left(x - \frac{6}{2} x\right)}^{2} + 14 + k = 0$

Discard the $x$ from $\frac{6}{2}$

$\text{ } 2 {\left(x - \frac{6}{2}\right)}^{2} + 14 + k = 0$
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The error comes from the $- \frac{6}{2}$ This is inside a bracket that is squared and then multiplied by the 2 outside the bracket

So the error is $2 \times {\left(- \frac{6}{2}\right)}^{2} = + \frac{36}{2} = + 18$

This has to be turned into 0 by $k$ so $\text{ } k + 18 = 0 \implies k = - 18$
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$\textcolor{b r o w n}{2 {\left(x - \frac{12}{2}\right)}^{2} + 14 + k = 0} \textcolor{g r e e n}{\text{->" } 2 {\left(x - \frac{6}{2}\right)}^{2} + 14 - 18 = 0}$

" "color(blue)(bar(underline(|" "2(x-3)^2-4=0" "|)))
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$\textcolor{b l u e}{\text{Determine x intercepts}}$

Write as:$\text{ } {\left(x - 3\right)}^{2} = \frac{4}{2}$

Square root both sides

$\text{ } x - 3 = \pm \sqrt{2}$

$\text{ } x = + 3 \pm \sqrt{2}$

$\textcolor{b l u e}{\text{ "x~~4.414" and "x~~1.586" to 3 decimal places}}$ May 2, 2016

$x = \sqrt{2} + 3$ OR $x = - \sqrt{2} + 3$
$x = 4.414$ OR $x = 1.586$ ( 3 dec places)

#### Explanation:

Completing the square is based on the consistency of the answers to the square of a binomial.

${\left(x - 3\right)}^{2} = {x}^{2} - 6 x + 9$
${\left(x - 5\right)}^{2} = {x}^{2} - 10 x + 25$
${\left(x + 6\right)}^{2} = {x}^{2} + 12 x + 36$
In all of the products above, $a {x}^{2} + b x + c$ we see the following:

$a = 1$
The first and last terms, $a \mathmr{and} c$ are perfect squares.'
There is a specific relationship between 'b' - the coefficient of the $x$ term and 'c'. Half of b, squared equals c.

Knowing this, it is always possible to add in a missing value for $c$ (ie complete the square) to make the square of a binomial, which can then be written as (x+?)^2

In $2 {x}^{2} - 12 x = - 14$, -14 has already been moved to the right hand side. The equation must be divided by 2 to make $a$ = 1

This gives:

${x}^{2} - 6 x = - 7$

NOW the correct value of $c$ can be added to BOTH sides of the equation to give the answer to the square of a binomial.

${x}^{2} - 6 x + \textcolor{red}{9}$ = -7 + color(red)(9) rArr [9 = (-6÷2)^2]
${\left(x - 3\right)}^{2} = 2 \Rightarrow$ where -3 is from -6÷2
$x - 3 = \pm \sqrt{2} \Rightarrow$ take the square root of both sides

This gives 2 possible answers for $x$.

$x = \sqrt{2} + 3$ OR $x = - \sqrt{2} + 3$