# Help with solving for x.

• November 18th 2007, 02:27 PM
Jeavus
Help with solving for x.
x^4 - 7 = 6x^2

I get this far:

x^4 - 7 = 6x^2
x^4 + 0x^3 - 6x^2 + 0x - 7 = 0

Factors of -7:
+/- 1, +/- 7

I don't know where to go from this point.

• November 18th 2007, 02:35 PM
Jhevon
Quote:

Originally Posted by Jeavus
x^4 - 7 = 6x^2

I get this far:

x^4 - 7 = 6x^2
x^4 + 0x^3 - 6x^2 + 0x - 7 = 0

Factors of -7:
+/- 1, +/- 7

I don't know where to go from this point.

the factor theorem is not needed here

bring everything to one side:

$x^4 - 6x^2 - 7 = 0$

$\Rightarrow \left( x^2 \right)^2 - 6 \left( x^2 \right) - 7 = 0$

this is a quadratic in $x^2$, it can be factored
• November 18th 2007, 02:36 PM
Jeavus
How can that be factored?

Ahh...alright. Let's see.

(x^2)^2 - 6(x^2) - 7 = 0
(x^2 - 7)(x^2+1) = 0

Is this correct?

How would I continue?
• November 18th 2007, 02:55 PM
Jhevon
Quote:

Originally Posted by Jeavus
How can that be factored?

Ahh...alright. Let's see.

(x^2)^2 - 6(x^2) - 7 = 0
(x^2 - 7)(x^2+1) = 0

Is this correct?

How would I continue?

how would you continue if it was a regular quadratic?