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

Question

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

1 Answer

Impact of this question

1944 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-12T04:00:24

This question doesn't have any real solutions. However it does have two complex solutions;
$x = \frac{1 \pm i \sqrt{39}}{4}$ $color(red)(x=(1+-isqrt(39))/4)$

Explanation:

$2 x^{2} - x = - 5$ $2x^2-x=-5$
$2 x^{2} - x + 5 = 0$ $2x^2-x+5=0$
$2 (x^{2} - \frac{x}{2} + \frac{5}{2}) = 0$ $2(x^2-x/2+5/2)=0$
$x^{2} - \frac{x}{2} + \frac{5}{2} = 0$ $x^2-x/2+5/2=0$
${(x - \frac{1}{4})}^{2} - \frac{1}{16} + \frac{5}{2} = 0$ $(x-1/4)^2-1/16+5/2=0$
${(x - \frac{1}{4})}^{2} + \frac{39}{16} = 0$ $(x-1/4)^2+39/16=0$
${(x - \frac{1}{4})}^{2} = - \frac{39}{16}$ $(x-1/4)^2=-39/16$
$x - \frac{1}{4} = \pm \sqrt{- \frac{39}{16}}$ $x-1/4=+-sqrt(-39/16)$
$x - \frac{1}{4} = \pm \frac{\sqrt{- 39}}{4} = \pm \frac{\sqrt{39} \times \sqrt{- 1}}{4}$ $x-1/4=+-sqrt(-39)/4=+-(sqrt(39)xxsqrt(-1))/4$
$x = \frac{1}{4} \pm \frac{\sqrt{39} \times \sqrt{- 1}}{4} = \frac{1}{4} \pm \frac{\sqrt{39} \times i}{4}$ $x=1/4+-(sqrt(39)xxsqrt(-1))/4=1/4+-(sqrt(39)xxi)/4$
$x = \frac{1 \pm i \sqrt{39}}{4}$ $color(red)(x=(1+-isqrt(39))/4)$

I hope it helps :)

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 2x2−x=−52x^2-x=-5 by completing the square?

1 Answer

Explanation:

Related questions

Impact of this question

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