# How do you solve 8x^2 + 14x + 5 = 0 by completing the square?

May 8, 2018

#### Answer:

$x = - \frac{5}{4} \text{ or } x = - \frac{1}{2}$

#### Explanation:

$\text{using the method of "color(blue)"completing the square}$

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

$\text{factor out 8}$

$\Rightarrow 8 \left({x}^{2} + \frac{7}{4} x + \frac{5}{8}\right) = 0$

• "add/subtract "(1/2"coefficient of the x-term")^2" to"
${x}^{2} + \frac{7}{4} x$

$8 \left({x}^{2} + 2 \left(\frac{7}{8}\right) x \textcolor{red}{+ \frac{49}{64}} \textcolor{red}{- \frac{49}{64}} + \frac{5}{8}\right) = 0$

$\Rightarrow 8 {\left(x + \frac{7}{8}\right)}^{2} + 8 \left(- \frac{49}{64} + \frac{5}{8}\right) = 0$

$\Rightarrow 8 {\left(x + \frac{7}{8}\right)}^{2} - \frac{9}{8} = 0$

$\Rightarrow 8 {\left(x + \frac{7}{8}\right)}^{2} = \frac{9}{8}$

$\text{divide both sides by 8}$

$\Rightarrow {\left(x + \frac{7}{8}\right)}^{2} = \frac{9}{64}$

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

$\sqrt{{\left(x + \frac{7}{8}\right)}^{2}} = \pm \sqrt{\frac{9}{64}} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x + \frac{7}{8} = \pm \frac{3}{8}$

$\text{subtract "7/8" from both sides}$

$\Rightarrow x = - \frac{7}{8} \pm \frac{3}{8}$

$\Rightarrow x = - \frac{7}{8} - \frac{3}{8} = - \frac{10}{8} = - \frac{5}{4}$

$\text{or } x = - \frac{7}{8} + \frac{3}{8} = - \frac{4}{8} = - \frac{1}{2}$

May 8, 2018

#### Answer:

$x = - \frac{10}{8} = - \frac{5}{4}$
$x = - \frac{4}{8} = - \frac{1}{2}$

#### Explanation:

Given: $\textcolor{g r e e n}{y = 0 = 8 {x}^{2} + 14 x + 5}$

Write as:
$\textcolor{g r e e n}{0 = 8 \left({x}^{2} + \frac{14}{8} x\right) + 5 \textcolor{w h i t e}{\text{ddd")->color(white)("ddd}} 0 = 8 \left({x}^{2} + \frac{7}{4} x\right) + 5}$

The 'perfect' square related to ${x}^{2} + \frac{7}{4} x \text{ is } {\left(x + \frac{7}{8}\right)}^{2}$

but $8 {\left(x + \frac{7}{8}\right)}^{2}$ introduces the value of $\textcolor{red}{8 \times {\left(\frac{7}{8}\right)}^{2}}$ that is not in the original equation. We put it in so we have to take it out.

$\textcolor{g r e e n}{0 = 8 \left({x}^{2} + \frac{7}{4} x\right) + 5}$

$\textcolor{g r e e n}{0 = 8 {\left(x + \frac{7}{8}\right)}^{2} \textcolor{red}{- \left[8 \times {\left(\frac{7}{8}\right)}^{2}\right]} + 5}$

$\textcolor{g r e e n}{0 = 8 {\left(x + \frac{7}{8}\right)}^{2} - \frac{9}{8}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Check}}$

$8 {\left(x + \frac{7}{8}\right)}^{2} - \frac{9}{8}$
$8 \left({x}^{2} + \frac{7}{4} x + \frac{49}{64}\right) - \frac{9}{8}$

$8 {x}^{2} + 14 x + \frac{49}{8} - \frac{9}{8}$

$8 {x}^{2} + 14 x + \frac{40}{8}$

$8 {x}^{2} + 14 x + 5$ as required!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(green)(0=8(x+7/8)^2-9/8color(white)("dd") ->color(white)("dd")+9/8=8(x+7/8)^2

$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{ddddddddddddddddddd")->color(white)("dddd}} \frac{9}{64} = {\left(x + \frac{7}{8}\right)}^{2}}$

$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{ddddddddddddddddddd")->color(white)("d.d")+-3/8=color(white)("d}} x + \frac{7}{8}}$

$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{ddddddddddddddddddd")->color(white)("ddd.dd}} x = - \frac{7}{8} \pm \frac{3}{8}}$

$x = - \frac{10}{8} = - \frac{5}{4}$
$x = - \frac{4}{8} = - \frac{1}{2}$ May 8, 2018

#### Answer:

$x = - \frac{1}{2} \mathmr{and} x = - \frac{5}{4}$

#### Explanation:

We have,

$8 {x}^{2} + 14 x + 5 = 0$

$\implies 8 {x}^{2} + 14 x = - 5$

$\implies 8 {x}^{2} + 14 x + K = K - 5. . . \to \left(A\right)$

Now we have to find $K$,such that,$\left(8 {x}^{2} + 14 x + K\right)$is a square.

Here,

${I}^{s t} t e r m = 8 {x}^{2}$

$I {I}^{n d} t e r m = 14 x$

$I I {I}^{r d} t e r m = K$

Formula for color(red)(III^(rd)term=(II^(nd)term)^2/(4xxI^(st)term)...to(psi)

$i . e . K = {\left(14 x\right)}^{2} / \left(4 \times 8 {x}^{2}\right) = \frac{196 {x}^{2}}{32 {x}^{2}} = \frac{49}{8}$

So,from$\left(A\right)$,we get

$8 {x}^{2} + 14 x + \frac{49}{8} = \frac{49}{8} - 5$

${\left(\sqrt{8} x\right)}^{2} + 2 \left(\sqrt{8} x\right) \left(\frac{7}{\sqrt{8}}\right) + {\left(\frac{7}{\sqrt{8}}\right)}^{2} = \frac{9}{8}$

${\left(\sqrt{8} x + \frac{7}{\sqrt{8}}\right)}^{2} = {\left(\frac{3}{\sqrt{8}}\right)}^{2}$

$\implies \sqrt{8} x + \frac{7}{\sqrt{8}} = \pm \frac{3}{\sqrt{8}}$

$\implies 8 x + 7 = \pm 3$

$\implies 8 x + 7 = 3 \mathmr{and} 8 x + 7 = - 3$

$\implies 8 x = 3 - 7 \mathmr{and} 8 x = - 3 - 7$

$\implies 8 x = - 4 \mathmr{and} 8 x = - 10$

$\implies x = - \frac{1}{2} \mathmr{and} x = - \frac{5}{4}$

$\implies x = - 0.5 \mathmr{and} x = - 1.25$

Note: Formula color(red)((psi) for color(red)(III^(rd)term, is always true.