Question

How do you solve `2x^2-5x-3=0` by completing the square?

Answer 1

`x_1=3`, `x_2=-1/2`

Explanation:

Completing the square means that you want to write `2x^2-5x-3=0`
on the form `(x-a)^2=b`

First, let's get rid of the constant in front of the 2nd degree term:
`x^2-5/2x-3/2=0`
We want this on the form
`(x-a)^2=b`
i.e. `x^2-2ax+a^2=b`
Comparing term for term, we see that
`2a=5/2`, i.e. `a=5/4`
`(x-5/4)^2=x^2-5/2x+25/16`
We, therefore, have
`(x^2-5/2x+25/16)-25/16-3/2=0`
or `(x^2-5/2x+25/16)=25/16+24/16=49/16`
`(x-5/4)^2=(7/4)^2`
`x-5/4=+-7/4`
`x=5/4+-7/4`
Therefore `x_1= 5/4+7/4=3`
`x_2=5/4-7/4=-1/2`

Answer 2

`x = 3` or `x=-1/2`

Explanation:

Given:

`2x^2 -5x -3 =0`

On dividing this equatin by `2`, we get:

`x^2 -5/2x -3/2 =0`

`x^2 -5/2x = 3/2`

Add `(5/4)^2` on both sides

`x^2 -5/2x +(5/4)^2 = 3/2 + (5/4)^2`

[because `a^2 + b^2 + 2ab = (a+b)^2` ]

So

`(x-5/4)^2 = 3/2 + 25/16`

`(x-5/4)^2 = 49/16`

Taking the square root on both sides and solving it.

`x-5/4 = 7/4" "` or `" "x-5/4 = -7/4`

`x = 7/4+5/4" "` or `" "x = -7/4+5/4`

`x = 12/4" "` or `" "x = -2/4`

`x = 3" "` or `" "x = -1/2`