Question

How do you solve using the quadratic formula `2x^2 - 2x = 1`?

Answer 1

`x_1=(1+sqrt(3))/2 or x_2=(1-sqrt(3))/2`

Explanation:

`2x^2-2x=1|-1`
`2x^2-2x-1=0|:2`
`x^2-x-1/2=0`
`x_(1,2)=-(p/2)+-sqrt((p/2)^2-q)`
`x_(1,2)=-(-1/2)+-sqrt((-1/2)^2-(-1/2))`
`x_(1,2)=1/2+-sqrt(1/4+1/2)=1/2+-sqrt(3)/2`

Answer 2

`x=(1pmsqrt3)/2`

Explanation:

Minus `1`:

`2x^2-2x-1=0`

Use the formula:

`rArr x=-bpm(sqrt(b^2-4ac))/(2a)`

Use the quadratic to find the values:

`2x^2-2x-1`

`a=2`

`b=-2`

`c=-1`

`therefore` `rArr x=-(-2)pm(sqrt((-2)^2-4(2)(-1)))/(2(2))`

`rArr x=(1pmsqrt(3))/(2)`

`rArr (1+sqrt3)/2, and (1-sqrt3)/2`

You could change these to decimals or leave it in surd form, however, I will leave in surd form.