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#

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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.