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

1 Answer
Shwetank Mauria
May 26, 2016

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 andbeta 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.

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