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(6x^2-14x+12)/(x-1)^4 =0 can someone find me how much x is?

Guest Mar 4, 2017
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 #1
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Solve for x:
(6 x^2 - 14 x + 12)/(x - 1)^4 = 0

Multiply both sides by (x - 1)^4:
6 x^2 - 14 x + 12 = 0

Divide both sides by 6:
x^2 - (7 x)/3 + 2 = 0

Subtract 2 from both sides:
x^2 - (7 x)/3 = -2

Add 49/36 to both sides:
x^2 - (7 x)/3 + 49/36 = -23/36

Write the left hand side as a square:
(x - 7/6)^2 = -23/36

Take the square root of both sides:
x - 7/6 = (i sqrt(23))/6 or x - 7/6 = -(i sqrt(23))/6

Add 7/6 to both sides:
x = 7/6 + (0 + i/6) sqrt(23) or x - 7/6 = -(i sqrt(23))/6

Add 7/6 to both sides:
Answer: |x = 7/6 + (0 + i/6) sqrt(23)      or      x = 7/6 + (0 - i/6) sqrt(23)

Guest Mar 4, 2017
 #2
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(6x^2-14x+12)/(x-1)^4 =0 can someone find me how much x is?

 

\(\frac{(6x^2-14x+12)}{(x-1)^4} =0\\ \text{First thing to note is that x cannot be 1}\\ 6x^2-14x+12=0\\ 3x^2-7x+6=0\\ \triangle=49-4*3*6=49-72=-23\\ \text{Since the discriminant is negative there are no real solutions}\)

 

Guest #1 has found the complex (imaginary) roots for you but if you don't know what that means then the answer is

"no real solution"

Melody  Mar 4, 2017

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