Algebra

**OzzMan** · February 8th 2008, 01:34 PM

Solve for $m$
$5m^4+24m^2-36=0\;m^2=y\Rightarrow 5y^2+24y-36=0$
Then i plugged it into the quadratic formula
$y=\frac{-24\pm\sqrt{24^2-(4)(5)(-36)}}{10}\Rightarrow y=\frac{-24\pm36}{10}\;y=\frac{6}{5}\;y=-6$
From there i did $m^2=\frac{6}{5}\Rightarrow m=\pm\sqrt\frac{6}{5}$
$m^2=-6$ Which gives an imaginary answer, so I really cant use that as answer right?

P.S. I used $m=\pm\sqrt\frac{6}{5}$ as my answer and I got marked off 1 point anyone know why?

**galactus** · February 8th 2008, 01:42 PM

That's good, subbing y for m^2. Therefore, you get a quadratic.

Just factor it.

$5y^{2}+24y-36=0$

$(y+6)(5y-6)=0$

Now, sub m^2 back in:

$(m^{2}+6)(5m^{2}-6)=0$

Now, The M^2+6 results in 2 non-real solutions.

The 5m^2-6 will give two real solutions.

**OzzMan** · February 8th 2008, 01:47 PM

Do you think I should put the non real answers as well as the real answers on my test as the answers for what m equals?

**Soroban** · February 8th 2008, 01:49 PM

Hello, OzzMan!

Solve for $m$

$5m^4+24m^2-36\:=\:0$

Let $m^2\,=\,y\quad\Rightarrow\quad 5y^2+24y-36\:=\:0$

Then i plugged it into the quadratic formula:

$y\:=\:\frac{-24\pm\sqrt{24^2-(4)(5)(-36)}}{10}\quad\Rightarrow\quad y\:=\:\frac{-24\pm36}{10} \;=\;\frac{6}{5}\;-6$

From there i did $m^2\:=\:\frac{6}{5}\quad\Rightarrow\quad m\:=\:\pm\sqrt\frac{6}{5}$

$m^2\:=\:-6$ which gives an imaginary answer.

P.S. I used $m\:=\:\pm\sqrt\frac{6}{5}$ as my answer and I got marked off 1 point.
Anyone know why?

Just a guess . . . maybe you were expected to rationalize that square root?

. . $\pm\sqrt{\frac{6}{5}} \;=\;\pm\sqrt{\frac{6}{5}\cdot\frac{5}{5}} \;=\;\pm\sqrt{\frac{30}{25}} \;=\;\pm\frac{\sqrt{30}}{\sqrt{25}} \;=\;\pm\frac{\sqrt{30}}{5}$

Oh yes, your reasoning and your work is correct . . . Nice going!

**OzzMan** · February 8th 2008, 01:53 PM

oh that makes a lot of sense thanks

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