How do you find the roots, real and imaginary, of #y=-5x^2 +(x -4)^2 # using the quadratic formula?

1 Answer
May 26, 2016

Answer:

We have two real zeros, which are irrational. These are #-1+-sqrt5#

Explanation:

Quadratic formula uses general form of quadratic equation #ax^2+bx+c=0# abd gives the solution of this equation as #x=(-b+-sqrt(b^2-4ac))/(2a)#. If #alpha# and#beta# are roots of this equation, then they are also zeros of the function #y=ax^2+bx+c#.

Hence, we should first convert the function #y=-5x^2+(x-4)^2#, to general form. Simplifying it we get,

#y=-5x^2+x^2-8x+16#

or #y=-4x^2-8x+16# and hence zeros of this function are

#x=(-(-8)+-sqrt((-8)^2-4*(-4)*16))/(2*(-4))#

or #x=(8+-sqrt(64+256))/(-8)=(8+-sqrt(320))/(-8)#

or #x=(8+-8sqrt5)/(-8)=-1+-sqrt5#

Hence, we have two real zeros, which are irrational.