How do you find the roots, real and imaginary, of #y=-(x-3)^2-(2x-3)^2-10 # using the quadratic formula?

1 Answer
Jan 23, 2018

Answer:

#x=(9+isqrt59)/5,##(9-isqrt59)/5#

#x=(9+sqrt(59)i)/5,##(9-sqrt(59)i)/5#

Explanation:

Given:

#y=-(x-3)^2-(2x-3)^2-10#

Expand #(x-3)^2# to #x^2-6x+9#.

#y=-(x^2-6x+9)-(2x-3)^2-10#

Expand #(2x-3)^2# to #4x^2-12x+9#.

#y=-(x^2-6x+9)-(4x^2-12x+9)-10#

Distribute the #-1# through both expressions.

#y=-x^2+6x-9-4x^2+12x-9-10#

Collect like terms.

#y=(-x^2-4x^2)+(6x+12x)+(-9-9-10)#

Simplify.

#y=-5x^2+18x-28# is a quadratic equation in standard form:

#y=ax^2+bx+c#,

where:

#a=-5#, #b=+18#, and #c=-28#

Find the roots using the quadratic formula.

The roots are the values for #x# when #y=0#, so first substitute #0# for #y#.

#0=-5x^2+18x-28#

Quadratic formula

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

Plug in the known values.

#x=(-18+-sqrt(18^2-4*-5*-28))/(2*-5)#

Simplify.

#x=(-18+-sqrt(-236))/(-10)#

Prime factorize #-236#.

#x=(-18+-sqrt(-1xx2xx2xx59))/(-10)#

Simplify.

#x=(-18+-2isqrt59)/(-10)#

Two negatives make a positive.

#x=(18+-2isqrt59)/10#

Simplify by dividing by #2#.

#x=(color(red)cancel(color(black)(18))^9+-color(red)cancel(color(black)(2))^1isqrt59)/color(red)cancel(color(black)(10))^5#

#x=(9+-isqrt59)/5#

Solutions for #x# (the roots)

#x=(9+isqrt59)/5,##(9-isqrt59)/5#

They can also be written with the imaginary number #i# after the square root.

#x=(9+sqrt(59)i)/5,##(9-sqrt(59)i)/5#