Question 616490: (x^2-1)^2-14(x^2-1)+33=0
I need to find the solutions for this, but can not figure out how to solve it. Any help so I can learn this would be very much appreciated.
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! Notice that if I:
Let y = (x^2-1)
then I can rewrite:
(x^2-1)^2-14(x^2-1)+33=0
as:
y^2-14y+33=0
Now, it looks like a regular quadratic which we can factor as:
(y-3)(y-11) = 0
y = {3, 11}
.
But we don't what 'y' we want to know what 'x' is.
so set both to:
y = (x^2-1)
and solve for x:
.
3 = x^2-1
4 = x^2
+- 2 = x
.
11 = x^2-1
12 = x^2
sqrt(12) = x
2*sqrt(3) = x
.
x = {-2*sqrt(3), 2*sqrt(3), -2, 2}
|
|
|
