Question

How do you find the roots, real and imaginary, of `y=2(x+3)^2-(x-3)^2` using the quadratic formula?

Answer 1

`y=2(x+3)^2-(x-3)^2`
`implies y=2(x^2+6x+9)-(x^2-6x+9)`
`implies y=2x^2+12x+18-x^2+6x-9`
`implies y=x^2+18x+9`
For finding roots #`becomes`0#. #implies x^2+18x+9=0#

The quadratic formula is
If `ax^2+bx+c=0` then `x=(-b+-sqrt(b^2-4ac))/(2a)`
Here `a=1` `b=18` and `c=9`

`implies x=(-18+-sqrt(18^2-4*1*9))/(2*1)`

`implies x=(-18+-sqrt(324-36))/(2)`

`implies x=(-18+-sqrt(324-36))/(2)`

`implies x=(-18+-sqrt(288))/(2)`

`implies x=(-18+-sqrt(288))/(2)`

`implies x=(-18+-12sqrt(2))/(2)`

`implies x=(2(-9+-6sqrt(2)))/(2)`

`implies x=(-9+-6sqrt(2))`