How do you find the roots, real and imaginary, of #y=-9x^2 -22x-55 # using the quadratic formula?

1 Answer
May 24, 2017

Answer:

#x=-(11+isqrt(384))/9, -(11-isqrt(384))/9#

Refer to the process in the explanation.

Explanation:

Substitute #0# for #y#. The roots are the values for #x# when #y# equals #0#. Since the graph is a parabola, there will be two roots.

#-9x^2-22x-55=0# is a quadratic equation in standard form:

#ax^2+bx+c#, where #a=-9#, #b=-22#, #c=-55#.

Quadratic Formula

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Insert the values for #a, b, and c# into the formula.

#x=(-(-22)+sqrt((-22)^2-4*-9*-55))/(2*-9)#

Simplify.

#x=(22+-sqrt(484-1980))/-18#

Simplify.

#x=(22+-sqrt(-1496))/-18#

Prime factorize #-1496#

#x=(22+-sqrt(-1xx2xx2xx2xx11xx17))/-18#
http://www.calculatorsoup.com/calculators/math/prime-factors.php

Simplify.

#x=(22+-isqrt(-1xx2^2xx2xx11xx17))/-18#

Simplify.

#x=(22+-2isqrt(384))/-18#

Simplify. (Divide #22#, #2#, #18# by #2#.)

#x=-(11+isqrt(384))/9,##x=-(11-isqrt(384))/9#