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

$\displaystyle Let A=x^{2}$

$\displaystyle A^2+A-42=0$

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

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

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

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

2. ## Re: Quadratic in form

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

3. ## Re: Quadratic in form

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

4. ## Re: Quadratic in form

Originally Posted by theloser
Separate Question:
$\displaystyle x^{4}+6x^{2}-91=0$
I did the exact same thing but ended up with
$\displaystyle (A+3)^2=100$
$\displaystyle x^2=-3\pm10$
$\displaystyle x^4+6x^2-91 = (x^2+13)(x^2-7) = (x + i\sqrt{13})(x - i\sqrt{13})(x + \sqrt{7})(x - \sqrt{7})$

5. ## Re: Quadratic in form

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

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

6. ## Re: Quadratic in form

I see that it can factor but I hate to factor big numbers so I instead tried to complete the square on $\displaystyle A^2+6A-91=0$ can that be done?

7. ## Re: Quadratic in form

Originally Posted by theloser
i see that it can factor but i hate to factor big numbers so i instead tried to complete the square on $\displaystyle a^2+6a-91=0$ can that be done?
$\displaystyle (a+3)^2=100$

8. ## Re: Quadratic in form

But does this give me the correct answer?
After that it would be
A=13
A=7

9. ## Re: Quadratic in form

Originally Posted by theloser
But does this give me the correct answer?
After that it would be
A=13
A=7
NO INDEED!
It gives $\displaystyle A+3=\pm 10$ or $\displaystyle A=-13\text{ or }A=7$