How do you find the roots, real and imaginary, of y= -3x^2-4x+(2x- 1 )^2 using the quadratic formula?

2 Answers
Aug 14, 2017

4 +- [ (15)^(1/2) ] / 2

Explanation:

If you expand the right side you should get
x^2 - 8x +1
Then just plug in to
x = ( -b +- [b^2 - 4ac]^(1/2) ) / (2a)
where a = 1 b = -8 c = 1

Aug 14, 2017

x=4+sqrt15,4-sqrt15

Refer to the explanation for the process.

Explanation:

Given:

y=-3x^2-4x+(2x-1)^2

FOIL (2x-1)^

y=-3x^2-4x+[4x^2-4x+1]

Gather like terms.

y=(-3x^2+4x^2)-(4x-4x)+1

Combine like terms.

y=x^2-8x+1 larr standard form: y=ax^2+bx+c,

where:

a=1, b=-8, and c=1

Substitute 0 for y and solve for x using the quadratic equation.

0=x^2-8x+1

Quadratic Formula

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

Plug in the given values.

x=(-(-8)+-sqrt((-8)^2-4*1*1))/(2*1)

Simplify.

x=(8+-sqrt(64-4))/2

x=(8+-sqrt60)/2

Prime factorize 60.

x=(8+-sqrt(2xx2xx3xx15))/2

Simplify.

x=(8+-2sqrt15)/2

Simplify.

x=(color(red)cancel(color(black)(8^4))+-color(red)cancel(color(black)(2^1))sqrt15)/color(red)cancel(color(black)(2^1))

x=4+-sqrt15

Roots

x=4+sqrt15,4-sqrt15