How do you solve $2 a^{2} + 8 a - 5 = 0$ $2a^2 + 8a − 5 = 0$ by completing the square?

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How do you solve $2 a^{2} + 8 a - 5 = 0$ $2a^2 + 8a − 5 = 0$ by completing the square?

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Answer 1 · 2015-07-06T19:11:54

$a = \pm \sqrt{\frac{13}{2}} - 2$ $a =+-sqrt(13/2)-2$

Explanation:

First, we put the constant in the right-hand side :
$2 a^{2} + 8 a - 5 = 0$ $2a^2+8a-5=0$
$\Leftrightarrow 2 a^{2} + 8 a = 5$ $<=>2a^2+8a=5$

Then we want to transform $2 a^{2} + 8 a$ $2a^2+8a$ to something like ${(a + b)}^{2}$ $(a+b)^2$

Recall : ${(x + y)}^{2} = x^{2} + 2 \cdot x y + y^{2}$ $(color(blue)x+color(red)y)^2 = color(blue)(x^2) + color(green)2*color(blue)xcolor(red)y + color(red)(y^2)$

Here we have $2 a^{2} + 8 a$ $2a^2 + 8a$
$\Leftrightarrow 2 \times (a^{2} + 4 a)$ $<=> 2xx(color(blue)(a^2)+4a)$
$\Leftrightarrow 2 \times (a^{2} + 2 \cdot a \cdot 2)$ $<=> 2xx(color(blue)(a^2)+color(green)2*color(blue)a*color(red)2)$ (Then $y = 2$ $"y" = 2$ )
$\Leftrightarrow 2 \times (a^{2} + 2 \cdot a \cdot 2 + 2^{2} - 2^{2})$ $<=> 2xx(color(blue)(a^2)+color(green)2*color(blue)a*color(red)2+color(red)(2^2)-2^2)$ (We added and subtract "y^2" for factorize later)

$\Leftrightarrow 2 \times ({(a + 2)}^{2} - 4)$ $<=> 2xx((a+2)^2-4)$

Put the last expression inside the equality :

$2 a^{2} + 8 a = 5$ $2a^2+8a=5$

$\Leftrightarrow 2 \times ({(a + 2)}^{2} - 4) = 5$ $<=> 2xx((a+2)^2-4)=5$

$\Leftrightarrow 2 \times {(a + 2)}^{2} - 8 = 5$ $<=> 2xx(a+2)^2 - 8 = 5$

$\Leftrightarrow 2 \times {(a + 2)}^{2} = 13$ $<=> 2xx(a+2)^2 = 13$

$\Leftrightarrow {(a + 2)}^{2} = \frac{13}{2}$ $<=>(a+2)^2 = 13/2$

$\Leftrightarrow a + 2 = \sqrt{\frac{13}{2}} or a + 2 = - \sqrt{\frac{13}{2}}$ $<=>a+2 = sqrt(13/2) or a+2 = -sqrt(13/2)$

Then $a = \pm \sqrt{\frac{13}{2}} - 2$ $a =+-sqrt(13/2)-2$

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How do you solve $2 a^{2} + 8 a - 5 = 0$ $2a^2 + 8a − 5 = 0$ by completing the square?

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