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

1 Answer
Meave60
Oct 23, 2017

Roots

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

Explanation:

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

Quadratic formula

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

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