How do you solve using the completing the square method 3x^2-67=6x+55?

Feb 3, 2018

$x = 1 \pm \frac{5 \sqrt{15}}{3}$

Explanation:

$\text{rearrange into standard form}$

•color(white)(x)ax^2+bx+c=0

$\Rightarrow 3 {x}^{2} - 6 x - 122 = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{to "color(blue)"complete the square}$

• " coefficient of "x^2" term must be 1"

$\Rightarrow 3 \left({x}^{2} - 2 x - \frac{122}{3}\right) = 0$

• " add/subtract "(1/2"coefficient of x-term")^2" to"
${x}^{2} - 2 x$

$3 \left({x}^{2} + 2 \left(- 1\right) x \textcolor{red}{+ 1} \textcolor{red}{- 1} - \frac{122}{3}\right) = 0$

$\Rightarrow 3 {\left(x - 1\right)}^{2} + 3 \left(- 1 - \frac{122}{3}\right) = 0$

$\Rightarrow 3 {\left(x - 1\right)}^{2} - 125 = 0$

$\Rightarrow 3 {\left(x - 1\right)}^{2} = 125$

$\Rightarrow {\left(x - 1\right)}^{2} = \frac{125}{3}$

$\textcolor{b l u e}{\text{take square root of both sides}}$

$\Rightarrow x - 1 = \pm \sqrt{\frac{125}{3}} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x = 1 \pm \frac{5 \sqrt{5}}{\sqrt{3}}$

$\Rightarrow x = 1 \pm \frac{5 \sqrt{15}}{3} \leftarrow \textcolor{b l u e}{\text{rationalise denominator}}$

$\Rightarrow x \approx - 5.45 \text{ or "x~~7.45" to 2 dec. places}$

Feb 3, 2018

$x = 1 \pm \frac{5 \sqrt{15}}{3}$

Explanation:

Given: $3 {x}^{2} - 67 = 6 x + 55$

By manipulating this we obtain

$3 {x}^{2} - 6 x - 122 = 0$

Set this as $y = 0 = 3 {x}^{2} - 6 x - 122$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Completing the square}}$

Consider the perfect square format of

$y = {\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

Our objective is to manipulate $y = 3 {x}^{2} - 6 x - 122$ to get it as close as we can to the perfect square format. This process introduces an error. This error is compensated for by including an 'adjustment' that takes the final result back to the value it should be. This correction I will show as $+ k$

Write as $y = 3 \left({x}^{2} - \frac{6}{3} x\right) - 122$ do not need $k$ yet

Now we will

$y = 3 {\left({x}^{\cancel{\textcolor{w h i t e}{.} 2}} - \frac{6}{2 \times 3} \cancel{x}\right)}^{2} - 122 + k$

We have halved the $- \frac{6}{3}$ and moved the squared from ${x}^{2}$ to outside the brackets. Also we have removed the $x$ from $- \frac{6}{3} x$

$y = 3 {\left(x \textcolor{c y a n}{- 1}\right)}^{2} - 122 \textcolor{red}{+ k}$

We need to set $\textcolor{w h i t e}{\text{d}} 3 \textcolor{c y a n}{{\left(- 1\right)}^{2}} \textcolor{red}{+ k} = 0$
$\implies \textcolor{red}{k} = - 3$ giving

$y = 3 {\left(x - 1\right)}^{2} - 122 + k \textcolor{w h i t e}{\text{dddd") ->color(white)("dddd}} y = 3 {\left(x - 1\right)}^{2} - 125$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Check}}$

$y = 3 {\left(x - 1\right)}^{2} - 123 \textcolor{w h i t e}{\text{ddd")->color(white)("ddd}} y = 3 \left({x}^{2} - 2 x + 1\right) - 125$

$\textcolor{w h i t e}{\text{ddddddddddddddddddd")->color(white)("ddd")y=3x^2-6x-122color(red)(larr" True}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determining the values that satisfy the equation}}$

$y = 0 = 3 {\left(x - 1\right)}^{2} - 125$

Add $125$ to both sides

$3 {\left(x - 1\right)}^{2} = 125$

Divide both sides by 3

${\left(x - 1\right)}^{2} = \frac{125}{3}$

Square root both sides

$x - 1 = \pm \sqrt{\frac{125}{3}}$

$x = + 1 \pm \sqrt{\frac{125}{3}}$

$125 = 25 \times 5 \to {5}^{2} \times 5$ giving:

$x = + 1 \pm \frac{5 \sqrt{5}}{\sqrt{3}}$

Mathematically it is considered that to have a root in the denominator is not good practice. So lets get rid of it.

Multiply the roots by 1 but in the form of $1 = \frac{\sqrt{3}}{\sqrt{3}}$

$x = 1 \pm \left[\frac{5 \sqrt{5}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\right]$

$x = 1 \pm \frac{5 \sqrt{15}}{3}$