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

2 Answers
Sep 16, 2016

Answer:

Two imaginary roots.
#x = (1 +- isqrt47)/8#

Explanation:

Develop and bring the equation to standard form:
#4(-x - 1)^2 = 4[-(x + 1)]^2 = 4(x + 1)^2 = 4(x^2 + 2x + 1) = 4x^2 + 8x + 4#.
#y = 12x^2 - 12x + 8 + (4x^2 + 8x + 4) = 0#
#y = 16x^2 - 4x + 12 = 4(4x^2 - x + 3) = 0#
Solve the quadratic equation by the improved quadratic formula (Socratic Search).
#y = 4x^2 - x + 3 = 0#
#D = 1 - 48 = -47 < 0# --> #d = +- isqrt47#
Since D < 0, there are 2 imaginary roots.
#x = -b/(2a) +- d/(2a) = 1/8 +- (isqrt47)/8 = (1 +- isqrt47)/8#

Sep 16, 2016

Answer:

#x=1/8+-sqrt(47)/8color(white)(.)i#

Explanation:

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

Squaring the bracket

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

giving:

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

#y=16x^2-4x+12 color(red)(larr" slightly different approach hear.")#

Set #y=0#

#=>0=16x^2-4x+12#

divide both sides by 4

Note that #0/4=0#

#=>0=4x^2-x+3#

Compare to #y=ax^2+bx+c " "->" "x=(-b+-sqrt(b^2-4ac))/(2a)#

Where#" "a=4"; "b=-1"; "c=3#

#=>x=(1+-sqrt((-1)^2-4(4)(3)))/(2(4))#

#x=(1+-sqrt(1-48))/8#

#x=1/8+-sqrt(-47)/8#

But 47 is a prime number so we have

#x=1/8+-sqrt(47)/8color(white)(.)i#