How do you find the roots, real and imaginary, of #y= 12x^2 - 4x + 8- (-x-1)^2 # using the quadratic formula?

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Answer 1 · 2017-10-23T23:05:54

Roots

#x=(3+2isqrt17)/11,##(3-2isqrt17)/11#

Solve:

#y=12x^2-4x+8-(-x-1)^2#

First simplify the parentheses.

#y=12x^2-4x+8-(x^2+2x+1)#

Simplify.

#y=12x^2-4x+8-x^2-2x-1#

Gather like terms.

#y=(12x^2-x^2)+(-4x-2x)+(8-1)#

Combine like terms.

#y=11x^2-6x+7# is a quadratic equation in standard form:

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

where:

#a=11#, #b=-6#, and #c=7#

Substitute #0# for #y#.

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

Plug the known values into the formula.

#x=(-(-6)+-sqrt((-6)^2-4*11*7))/(2*11)#

Simplify.

#x=(6+-sqrt(36-308))/22#

Simplify.

#x=(6+-sqrt(-272))/22#

Prime factorize #272#.

#x=(6+-(-(2xx2)xx(2xx2)xx17))/22#

Simplify.

#x=(6+-4isqrt17)/22#

Reduce.

#x=[(6+-4isqrt17)/22]/2#

#x=(3+-2isqrt17)/11#

Roots

#x=(3+2isqrt17)/11,##(3-2isqrt17)/11#