How do you solve $2 x^{2} + 5 x - 1 = 0$ $2x^2+5x-1=0$ by completing the square?

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How do you solve $2 x^{2} + 5 x - 1 = 0$ $2x^2+5x-1=0$ by completing the square?

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Answer 1 · 2018-04-19T12:03:33

$2 {(x + 1.25)}^{2} - 4.125 = 0$ $2(x+1.25)^2-4.125=0$

Explanation:

We first take the first two terms and factor out the coefficient of $x^{2}$ $x^2$ :
$\frac{2 x^{2}}{2} + \frac{5 x}{2} = 2 (x^{2} + 2.5 x)$ $(2x^2)/2+(5x)/2=2(x^2+2.5x)$

Then we divide by $x$ $x$ , half the integer and square what remains:
$2 (\frac{x^{2}}{x} + 2.5 \frac{x}{x}) 2 = 2 (x + 2.5)$ $2(x^2/x+2.5x/x)2=2(x+2.5)$
$2 (x + 2, \frac{5}{2}) = 2 (x + 1.25)$ $2(x+2,5/2)=2(x+1.25)$
$2 {(x + 1 . .25)}^{2}$ $2(x+1..25)^2$

Expand out the bracket:
$2 x^{2} + 2.5 x + 2.5 x + 2 ({1.25}^{2}) = 2 x^{2} + 5 x + 3.125$ $2x^2+2.5x+2.5x+2(1.25^2)=2x^2+5x+3.125$

Make it equal the original equations:
$2 x^{2} + 5 x + 3.125 + a = 2 x^{2} + 5 x - 1$ $2x^2+5x+3.125+a=2x^2+5x-1$

Rearrange to find $a$ $a$ :
$a = - 1 - 3.125 = - 4.125$ $a=-1-3.125=-4.125$

Put in $a$ $a$ to the factorised equation:
$2 {(x + 1.25)}^{2} - 4.125 = 0$ $2(x+1.25)^2-4.125=0$

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How do you solve $2 x^{2} + 5 x - 1 = 0$ $2x^2+5x-1=0$ by completing the square?

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Explanation:

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