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
Nghi N.
Sep 16, 2016

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

Tony B
Sep 16, 2016

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

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