+0

# need help

0
86
2

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

Guest Mar 4, 2017
Sort:

#1
0

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

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

x = 7/6 + (0 + i/6) sqrt(23) or x - 7/6 = -(i sqrt(23))/6

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

Guest Mar 4, 2017
#2
+90622
+5

(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

### 17 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details