How do you solve by completing the square 2x^2+7x-1=0?

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How do you solve by completing the square 2x^2+7x-1=0?

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Answer 1 · 2015-04-01T02:07:16

$2 x^{2} + 7 x - 1 = 0$ $2x^2 + 7x - 1 = 0$

Remove the constant from the left-side by adding $1$ $1$ to both sides
$2 x^{2} + 7 x = 1$ $2x^2+7x = 1$

Divide all terms by $2$ $2$ to reduce the $x^{2}$ $x^2$ coefficient to $1$ $1$
$x^{2} + \frac{7}{2} x = \frac{1}{2}$ $x^2 + 7/2 x = 1/2$

If the first 2 terms of ${(x + a)}^{2}$ $(x+a)^2$ evaluate as $x^{2} + \frac{7}{2} x$ $x^2 + 7/2x$
then $a = \frac{7}{4}$ $a = 7/4$

Add (7/4)^2 to both sides of the equation
$x^{2} + \frac{7}{2} x + {(\frac{7}{4})}^{2} = \frac{1}{2} + \frac{49}{16}$ $x^2+7/2 x +(7/4)^2 = 1/2 + (49)/(16)$

Rewrite the left-side as a square (and simplify the right side)
${(x + \frac{7}{4})}^{2} = \frac{57}{16}$ $(x+7/4)^2 = (57)/(16)$

Take the square root of both sides (remember both plus and minus)
$x + \frac{7}{4} = \pm \frac{\sqrt{57}}{4}$ $x+7/4 = +-sqrt(57)/4$

$x = \frac{- 7 + \sqrt{57}}{4}$ $x = (-7 +sqrt(57))/4$
or
$x = \frac{- 7 - \sqrt{57}}{4}$ $x=(-7-sqrt(57))/4$

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How do you solve by completing the square 2x^2+7x-1=0?

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