Question #296de

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Answer 1 · 2018-01-08T21:08:07

$x \approx - 2.31, - 0.19$ $x~~-2.31,-0.19$

First we square both sides:
${(\frac{\sqrt{4}}{\sqrt{4 x + 9}})}^{2} = {(\sqrt{8 x + 2})}^{2}$ $(sqrt(4)/sqrt(4x+9))^2=(sqrt(8x+2))^2$

$\frac{{(\sqrt{4})}^{2}}{{(\sqrt{4 x + 9})}^{2}} = {(\sqrt{8 x + 2})}^{2}$ $(sqrt(4))^2/(sqrt(4x+9))^2=(sqrt(8x+2))^2$

$\frac{4}{4 x + 9} = 8 x + 2$ $4/(4x+9)=8x+2$

Now we can rearrange and make it eqial to 0:
$4 = (8 x + 2) (4 x + 9)$ $4=(8x+2)(4x+9)$
$(8 x + 2) (4 x + 9) - 4 = 0$ $(8x+2)(4x+9)-4=0$

Now expand brackets:

$(8 x \cdot 4 x) + (2 \cdot 4 x) + (8 x \cdot 9) + (2 \cdot 9) - 4 = 0$ $(8x*4x)+(2*4x)+(8x*9)+(2*9)-4=0$

$32 x^{2} + 8 x + 72 x + 18 - 4 = 0$ $32x^2+8x+72x+18-4=0$

$32 x^{2} + 80 x + 14 = 0$ $32x^2+80x+14=0$

We can remove out common factors:

$32 x^{2} + 80 x + 14 = 0$ $32x^2+80x+14=0$

$16 x^{2} + 40 x + 7 = 0$ $16x^2+40x+7=0$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{- 40 \pm \sqrt{40^{2} - 4 (16 \cdot 7)}}{32}$ $x=(-40+-sqrt(40^2-4(16*7)))/32$

$= \frac{- 40 \pm \sqrt{1600 - 4 (112)}}{32}$ $=(-40+-sqrt(1600-4(112)))/32$

$= \frac{- 40 \pm \sqrt{1600 - 448}}{32}$ $=(-40+-sqrt(1600-448))/32$

$= \frac{- 40 \pm \sqrt{1152}}{32}$ $=(-40+-sqrt(1152))/32$

$= \frac{- 40 + \sqrt{1152}}{32} or \frac{- 40 - \sqrt{1152}}{32}$ $=(-40+sqrt(1152))/32or(-40-sqrt(1152))/32$

$\approx - 2.31 or - 0.19$ $~~-2.31or-0.19$