Question

How do you find the roots, real and imaginary, of `y= 8x^2 - 10x + 14-(3x-1)^2` using the quadratic formula?

Answer 1

`x = -2 + sqrt 17, -2 - sqrt17`

Explanation:

`y = 8 x^2 - 10 x + 14 - (3 x -1)^2`

`y = 8 x^2 - 10 x + 14 - (9x^2 -6x +1)`

`y = 8 x^2 - 9 x^2 - 10 x + 6 x + 14 - 1`

`y = - x^2 - 4 x + 6 x + 13`

to find it root when `y = 0`
`0 = - x^2 - 4 x + 6 x + 13`
`a = -1, b =-4 and c = 13`

`x = (-b +- sqrt(b^2 -4ac))/(2a)`
`x = (-(-4) +- sqrt((-4)^2 -4(-1)(13)))/(2(-1))`

`x = ((4) +- sqrt((16 + 52)))/(-2)`

`x = (4 +- sqrt(68))/(-2) = (4 +- sqrt(4 * 17))/(-2) = (4 +- 2 sqrt( 17))/(-2`

`x = -[(2 +- sqrt(17)]`

`x = -2 + sqrt 17, -2 - sqrt17`