Thursday

December 18, 2014

December 18, 2014

Posted by **Anonymous** on Thursday, May 22, 2008 at 10:06pm.

Here's my work, but I just went around in circles.....

0 = x^4 - 6x^2 + 5

-5 = x^4 - 6x^2

-5 = x^2 (x^2 - 6)

-5 / x^2 = x^2 - 6

(-5 / x^2) + 6 = x^2

(-5 + 6x^2) / (x^2) = x^2

-5 + 6x^2 = x^2 * x^2

-5 + 6x^2 = x^4

0 - x^4 - 6x^2 + 5

- Math -
**Reiny**, Thursday, May 22, 2008 at 10:14pmNO, you have not factored anything at all

x^4 - 6x^2 + 5 = 0

(x^2 - 1)(x^2 - 5) = 0

(x-1)(x+1)x^2 - 5) = 0 if you factor over the rationals

(x-1)(x+1)x-√5)(x+√5) = 0 if you factor over the reals

- Math -
**JJ**, Saturday, February 9, 2013 at 1:11pmstart by treating the problem like it is a quadratic. Let u = x^2

u^2 - 6u + 5

Factors as: (u-5)(u-1)

(x^2 -5)(x^2 -1) =0

Whenever there is = 0, Normally, you have to solve for x and not stop at the factoring step.

Set each factor = to zero.

(x^2-5)=0 or (x^2 - 1) = 0

x^2 = 5 or x^2 = 1

x = square root of 5 divided by 2

x = minus the square root of 5 divided by 2

x = 1

x = -1

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