Question

How do you find the roots, real and imaginary, of `y=2x^2 + 13x + 6+4(x -1)^2` using the quadratic formula?

Answer 1

`x = (-5 + sqrt(-215))/12` or `x = (-5 - sqrt(-215))/12`

Explanation:

First expand and simplify to ensure just one term for each power of x.
`y = 2x^2 +13x +6 +4(x^2 -2x + 1)`
`y = 2x^2 +13x + 6 + 4x^2 -8x +4`
`y=6x^2 +5x +10`
Then use `x = (-b +- sqrt(b^2 - 4ac))/(2a)`
`x = (-5 +- sqrt( 5^2 - 4*6*10))/(2*6)`
`x = (-5 +-sqrt(25 - 240))/12`
`x = (-5 +- sqrt(-215))/12`
There are only imaginary roots to this equation, meaning that in graphical terms the graph never intersects the x axis.