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

1 Answer
George C.
Dec 3, 2017

x = 1/4+-sqrt(41)/4

Explanation:

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

A^2-B^2=(A-B)(A+B)

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

0 = 8(2x^2-x-5)

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

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

color(white)(0) = (4x-1)^2-(sqrt(41))^2

color(white)(0) = ((4x-1)-sqrt(41))((4x-1)+sqrt(41))

color(white)(0) = (4x-1-sqrt(41))(4x-1+sqrt(41))

So:

4x = 1+-sqrt(41)

and:

x = 1/4+-sqrt(41)/4

Note that multiplying by 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 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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