Quadratic in form

**theloser** · July 16th 2011, 10:32 AM

$x^{4}+1x^{2}-42=0$

$Let A=x^{2}$

$A^2+A-42=0$

$\left ( A+7 \right )\left ( A-6 \right )=0$

$x^{2}=-7$ $x^{2}=6$

$x=\pm \sqrt{-7} x=\pm \sqrt{6}$

I need help I don't think I did this correctly.

**Siron** · July 16th 2011, 10:45 AM

It's almost correct, $x=\pm \sqrt{6}$ is a solution of the equation but you have to reject $x=\pm \sqrt{-7}$ because $\sqrt{-7}$ is undefined in $\mathbb{R}$ .

**theloser** · July 16th 2011, 11:07 AM

Separate Question:
$x^{4}+6x^{2}-91=0$
I did the exact same thing but ended up with
$(A+3)^2=100$
$x^2=-3\pm10$

**skeeter** · July 16th 2011, 11:16 AM

Originally Posted by theloser

Separate Question:
$x^{4}+6x^{2}-91=0$
I did the exact same thing but ended up with
$(A+3)^2=100$
$x^2=-3\pm10$

$x^4+6x^2-91 = (x^2+13)(x^2-7) = (x + i\sqrt{13})(x - i\sqrt{13})(x + \sqrt{7})(x - \sqrt{7})$

**Siron** · July 16th 2011, 11:16 AM

That's not correct.
$x^4+6x^2-91=0$
Let $x^2=t$ so:
$t^2+6t-91=0$
$\Leftrightarrow (t-7)(t+13)=0$
$\Leftrightarrow x^2=7 \ \mbox{or} \ x^2=-13$
We can reject $x^2=-13$ in $\mathbb{R}$ and the solutions are now: $x=\pm \sqrt{7}$

When you have to solve it in $\mathbb{C}$ then:
$x= \pm i\sqrt{13}$ is also a solution.