Question

Find the roots of the following quadratic equation if they exist by the method of completing the square 2x²-7x+3=0?

Answer 1

See my proof below.

Explanation:

Completing the square:

`x^2-2*7/4x+49/16+3/2-49/16=0`
this is

`(x-7/4)^2-25/16=0`
using that

`(a^2-b^2=(a-b)(a+b)`
then we get

`(x-7/4+5/4)(x-7/4-5/4)=0`

`(x-1/2)(x-3)=0`

Answer 2

`x=1/2` and `x=3`

Explanation:

It is always good to write out the square as a reminder. We want the expression on the form:
`(x-a)^2=b`
`2(x-a)^2=2(x^2-2ax+a^2)=2x^2-7x=-3`
gives
`(x-a)^2=(x^2-2ax+a^2)=x^2-7/2x+(7/4)^2`
=`-3/2+(7/4)^2=(-24+49)/16=25/16`
where I have moved the constant over to the right side.

Comparing term for term we see that
`2a=7/2` or `a=7/4`
The square, therefore, is
`(x-7/4)^2=25/16`

We, therefore get:
=`x-7/4=+-5/4`

Therefore:
`x=7/4+-5/4`
This gives the two solutions `x=1/2` and `x=3`

Test:

We see the graph of `f(x)=2x^(2)-7x+3` crosses the x axis in x=0.5 and x=3