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

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

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Answer 1 · 2017-12-03T09:46:34

$x = \frac{1}{4} \pm \frac{\sqrt{41}}{4}$ $x = 1/4+-sqrt(41)/4$

Explanation:

Premultiply by $8$ $8$ (to avoid some fractions), complete the square and use the difference of squares identity:

$A^{2} - B^{2} = (A - B) (A + B)$ $A^2-B^2=(A-B)(A+B)$

with $A = (4 x - 1)$ $A=(4x-1)$ and $B = \sqrt{41}$ $B=sqrt(41)$ as follows:

$0 = 8 (2 x^{2} - x - 5)$ $0 = 8(2x^2-x-5)$

$0 = 16 x^{2} - 8 x - 40$ $color(white)(0) = 16x^2-8x-40$

$0 = {(4 x)}^{2} - 2 (4 x) + 1 - 41$ $color(white)(0) = (4x)^2-2(4x)+1-41$

$0 = {(4 x - 1)}^{2} - {(\sqrt{41})}^{2}$ $color(white)(0) = (4x-1)^2-(sqrt(41))^2$

$0 = ((4 x - 1) - \sqrt{41}) ((4 x - 1) + \sqrt{41})$ $color(white)(0) = ((4x-1)-sqrt(41))((4x-1)+sqrt(41))$

$0 = (4 x - 1 - \sqrt{41}) (4 x - 1 + \sqrt{41})$ $color(white)(0) = (4x-1-sqrt(41))(4x-1+sqrt(41))$

So:

$4 x = 1 \pm \sqrt{41}$ $4x = 1+-sqrt(41)$

and:

$x = \frac{1}{4} \pm \frac{\sqrt{41}}{4}$ $x = 1/4+-sqrt(41)/4$

Note that multiplying by $8$ $8$ has a couple of effects:

It makes the leading term into a perfect square ready for completing the square.
It makes the coefficient of $x$ $x$ into a multiple of twice the square root of the leading coefficient, thus avoiding having to use fractions until the end.

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

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Explanation:

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