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

Answer:

# 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

Answer:

#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#