How do you find the roots, real and imaginary, of $y= x^2 - 6x +3 -(2x- 3 )^2$ using the quadratic formula?

Question

$y= x^2 - 6x +3 -(2x- 3 )^2 =0$

1465 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-27T09:16:02

$=>D=-36 < 0=>$ the roots are imaginary numbers.

Here,

$y=x^2-6x+3-(2x-3)^2$

$y=x^2-6x+3-(4x^2-12x+9)$

$=>y=x^2-6x+3-4x^2+12x-9$

$=>y=-3x^2+6x-6$

If $y=0$ ,then $-3x^2+6x-6=0" is a quadratic equation ."$

Comparing with $ax^2+bx+c=0$ ,we get

$a=-3 ,b=6 and c=-6$

So, $"the "color(blue)"Discriminant"$ of the quadratic eqn. is :

$color(blue)(D=b^2-4ac)=6^2-4(-3)(-6)=36-72$

$=>D=-36 < 0=>$ the roots are imaginary numbers.

$i.e. x in CCto$ complex roots.