How do you find the roots, real and imaginary, of #y=3x^2 +4x+(x -4)^2 # using the quadratic formula?

1 Answer
Dec 15, 2015

Answer:

Convert the given expression into standard form and then apply the quadratic formula.
This should give you the (imaginary) roots #(1/2+isqrt(15)/2)# and #(1/2-isqrt(15)/2)#

Explanation:

The quadratic formula says that for a quadratic equation in the form:
#color(white)("XXX")y=color(red)(a)x^2+color(blue)(b)x+color(green)(c)#
the roots are given by
#color(white)("XXX")x=(-color(blue)(b)+-sqrt(color(blue)(b)^2-4(color(red)(a)color(green)(c))))/(2(color(red)(a))#

To apply this to the given equation:
#color(white)("XXX")y=3x^2+4x+(x-4)^2#
we must first convert the equation into standard form by expanding

#color(white)("XXX")y=3x^2+4x+ (x^2-8x+16)#

#color(white)("XXX")y=color(red)(4)x^2color(blue)(-4)x+color(green)(16)#

We can now apply the quadratic formula:
#color(white)("XXX")x=(-color(blue)(4)+-sqrt((color(blue)(-4))^2-4(color(red)(4))(color(green)(16))))/(2(color(red)(4)))#

#color(white)("XXXX")=(-cancel(4)+-cancel(4)sqrt(-15))/(2xxcancel(4))#

#color(white)("XXX")=1/2+-(isqrt(15))/2#