Question

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

Answer 1

`x_1=(1+sqrt(41))/4`
`x_2=(1-sqrt(41))/4`

Explanation:

From the given `2x^2-x-5=0`

Factor out the 2 from the first two terms so that the coefficient of `x^2` is 1

`2(x^2-1/2x)-5=0`

Now take note of the 1/2 coefficient of x. Divide this 1/2 by 2 it becomes 1/4. Square it, it will become 1/16. This 1/16 will be added and subtracted to the terms inside the grouping symbol.

Let us continue

`2(x^2-1/2x)-5=0`

`2(x^2-1/2x+1/16-1/16)-5=0`

Notice `x^2-1/2x+1/16` is a PST-Perfect Square Trinomial and
`x^2-1/2x+1/16=(x-1/4)^2`

so that

`2((x^2-1/2x+1/16)-1/16)-5=0`

`2((x-1/4)^2-1/16)-5=0`

Now, transpose the 5 to the right side then divide both sides by 2 then transpose the 1/16

`2((x-1/4)^2-1/16)-5=0`

`2((x-1/4)^2-1/16)=5`

`(cancel2((x-1/4)^2-1/16))/cancel2=5/2`

`(x-1/4)^2-1/16=5/2`

`(x-1/4)^2=5/2+1/16`

Simplify

`(x-1/4)^2=41/16`

Extract the square root of both sides

`sqrt((x-1/4)^2)=+-sqrt(41/16)`

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

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

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

`x_1=(1+sqrt(41))/4`
`x_2=(1-sqrt(41))/4`

God bless...I hope the explanation is useful.