Tough algebra

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February 11th 2010, 07:54 PM
Punch

Tough algebra

Solve for x in $\frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0$
February 11th 2010, 08:14 PM
Bacterius

OMG, did you try solving a cubic polynomial using the quadratic formula or what ? :o

Anyway, here's a start to simplification (I can't really do anything at the moment) : note that $\frac{\sqrt{a}}{a} = \frac{1}{\sqrt{a}}$ . Apply this to your equation with $a = x^2 + 3$
February 11th 2010, 08:18 PM
Punch

I didn't. It's just an equation which I got from differentiation and am required to solve it...
February 11th 2010, 08:26 PM
Bacterius

That was a mean curve, eh ...
February 11th 2010, 08:28 PM
tedii

$\frac{\sqrt{x^2+3}}{2\sqrt{1-x}}=\frac{x\sqrt{1-x}}{\sqrt{x^2+3}}$

Since the top only needs to equal zero solve for this equation. This will become a lot easier when you get the denominators to match.
February 11th 2010, 08:30 PM
Punch

Quote:

Originally Posted by tedii

$\frac{\sqrt{x^2+3}}{2\sqrt{1-x}}=\frac{x\sqrt{1-x}}{\sqrt{x^2+3}}$

Since the top only needs to equal zero solve for this equation. This will become a lot easier when you get the denominators to match.

Thanks, I think i got it, will try it now!
February 11th 2010, 08:37 PM
Punch

this was what I did, $x^2+3=2x(1-x)$

$x^2+3=2x-2x^2$

$3x^2-2x+3=0$

Unable to solve further...
February 11th 2010, 08:50 PM
tedii

Quote:

Originally Posted by Punch

Solve for x in $\frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0$

Looking at your derivative you made a mistake, I'm pretty sure. It should be

$\frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{2x}{\sqrt{x^2+3}})}{x^2+3}=0$

Try redoing everything with that.
February 11th 2010, 09:04 PM
Punch

Quote:

Originally Posted by Punch

Thanks, I think i got it, will try it now!

Quote:

Originally Posted by tedii

Looking at your derivative you made a mistake, I'm pretty sure. It should be

$\frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{2x}{\sqrt{x^2+3}})}{x^2+3}=0$

Try redoing everything with that.

No I have checked and indeed I am right
February 11th 2010, 09:46 PM
tedii

Quote:

Originally Posted by Punch

Solve for x in $\frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0$

You're right about the 2 not being right.

$\frac{\sqrt{x^2+3}(-\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0$

But maybe you forgot the chain-rule on the first part for the -x.

If not and you're sure your derivative is correct then your first equation has no maximums or minimums. And you don't need to find any zeros.
February 11th 2010, 09:50 PM
Soroban

Hello, Punch!

Quote:

$\text{Solve for }x\!: \;\;\frac{\sqrt{x^2+3}\left(\dfrac{1}{2\sqrt{1-x}}\right)-\sqrt{1-x}\left(\dfrac{x}{\sqrt{x^2+3}}\right)}{x^2+3}\;\; =\;\;0$

If this is a derivative, you lost a minus-sign . . .

I assume the original function is: / $f(x)\;=\;\frac{\sqrt{1-x}}{\sqrt{x^2+3}} \;=\;\frac{(1-x)^{\frac{1}{2}}} {(x^2+3)^{\frac{1}{2}}}$

Then: . $f'(x) \;=\;\frac{(x^2+3)^{\frac{1}{2}}\cdot\frac{1}{2}(1-x)^{-\frac{1}{2}}{\color{red}(-1)} - (1-x)^{\frac{1}{2}}\cdot\frac{1}{2}(x^2+3)^{-\frac{1}{2}}(2x)}{x^2+3}$

. . . . . . . . $=\;\frac{\sqrt{x^2+3}\left(\dfrac{{\color{red}-1}}{2\sqrt{1-x}}\right) - \sqrt{1-x}\left(\dfrac{x}{\sqrt{x^2+3}}\right)} {x^2+3}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Multiply top and bottom by $2\sqrt{x^2+3}\,\sqrt{1-x}$

. . and we have: . $\frac{-(x^2+3) - 2x(1-x)}{2(x^2+3)^{\frac{3}{2}}\sqrt{1-x}} \;=\;0$

. . which simplifies to: . $\frac{x^2-2x-3}{2(x^2+3)^{\frac{3}{2}}\sqrt{1-x}} \;=\;0$

And we have: . $x^2 - 2x - 3 \:=\:0 \quad\Rightarrow\quad (x+1)(x-3) \:=\:0$

. . Hence: . $x \;=\;-1,\;3$

But because of the $\sqrt{1-x}$ in the function, $x \leq 1$

. . Therefore, the only critical value is: . $x \;=\;-1$
February 12th 2010, 01:46 AM
Punch

Quote:

Originally Posted by Soroban

Hello, Punch!

If this is a derivative, you lost a minus-sign . . .

I assume the original function is: / $f(x)\;=\;\frac{\sqrt{1-x}}{\sqrt{x^2+3}} \;=\;\frac{(1-x)^{\frac{1}{2}}} {(x^2+3)^{\frac{1}{2}}}$

Then: . $f'(x) \;=\;\frac{(x^2+3)^{\frac{1}{2}}\cdot\frac{1}{2}(1-x)^{-\frac{1}{2}}{\color{red}(-1)} - (1-x)^{\frac{1}{2}}\cdot\frac{1}{2}(x^2+3)^{-\frac{1}{2}}(2x)}{x^2+3}$

. . . . . . . . $=\;\frac{\sqrt{x^2+3}\left(\dfrac{{\color{red}-1}}{2\sqrt{1-x}}\right) - \sqrt{1-x}\left(\dfrac{x}{\sqrt{x^2+3}}\right)} {x^2+3}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Multiply top and bottom by $2\sqrt{x^2+3}\,\sqrt{1-x}$

. . and we have: . $\frac{-(x^2+3) - 2x(1-x)}{2(x^2+3)^{\frac{3}{2}}\sqrt{1-x}} \;=\;0$

. . which simplifies to: . $\frac{x^2-2x-3}{2(x^2+3)^{\frac{3}{2}}\sqrt{1-x}} \;=\;0$

And we have: . $x^2 - 2x - 3 \:=\:0 \quad\Rightarrow\quad (x+1)(x-3) \:=\:0$

. . Hence: . $x \;=\;-1,\;3$

But because of the $\sqrt{1-x}$ in the function, $x \leq 1$

. . Therefore, the only critical value is: . $x \;=\;-1$

Sorry Soroban, this is just how careless I am... thanks