Solving radical equations x√3+x√2=1

[nek9876]

I'm trying to help my son with homework and can't get the following problem

$$x\sqrt{3}+x\sqrt{2}=1$$

Please help

 

[topsquark]

I'm trying to help my son with homework and can't get the following problem

$$x\sqrt{3}+x\sqrt{2}=1$$

Please help

Start by factoring the x from the left hand side:
[math]x \sqrt{3} + x \sqrt{2} = x \left ( \sqrt{3} + \sqrt{2} \right ) = 1[/math]

Can you finish from here?

-Dan

 

[MarkFL]

I'm trying to help my son with homework and can't get the following problem

$$x\sqrt{3}+x\sqrt{2}=1$$

Please help

Hello, and welcome to MHB! (Wave)

I was going to post essentially the same thing as Dan, but was beaten to the punch. :)

 

[nek9876]

Thanks for the help.

Ok so we had thought about factoring out the X and then we thought the next step would be to square both sides.

If we do that the one side is obviously 1, but then does the other side become 5xe^{2}?

 

[MarkFL]

Thanks for the help.

Ok so we had thought about factoring out the X and then we thought the next step would be to square both sides.

If we do that the one side is obviously 1, but then does the other side become 5xe^{2}?

No, but this is such a common error made by students that it has been named "The Freshman's Dream." In general we have:

$$\left(\sqrt{a}+\sqrt{b}\right)^2\ne a+b$$

What is true though is:

$$\left(\sqrt{a}+\sqrt{b}\right)^2=a+2\sqrt{ab}+b$$

Let's go back to:

$$x\left(\sqrt{3}+\sqrt{2}\right)=1$$

We wish to isolate $x$ on one side of the equation, and what stands in our way at the moment is that $x$ is being multiplied by the number $\sqrt{3}+\sqrt{2}$...so, we need to divide both sides by this value...in doing so, what do we have?

 

[nek9876]

That makes sense and I see the error we made in squaring it out.

So based on dividing both sides we would then isolate the x and have

x=1/(\sqrt{2}+\sqrt{3})

So then we do have x isolated, I would think at this point that this is the final answer

Also one question when typing this in, I must be using the wrong symbols on the side to hit square root, how do
I put it in the correct form so it looks better in the future?

Again thanks for the help

 

[MarkFL]

That makes sense and I see the error we made in squaring it out.

So based on dividing both sides we would then isolate the x and have

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

So then we do have x isolated, I would think at this point that this is the final answer

Yes, that would be acceptable, except in the case where you are instructed to rationalize the denominator where you could write:

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

Also one question when typing this in, I must be using the wrong symbols on the side to hit square root, how do
I put it in the correct form so it looks better in the future?

Again thanks for the help

In order to get $\LaTeX$ to render, you need to wrap the code in tags, the easiest of which is to click the $\sum$ button on the editor toolbar, which will generate $$$$ tags. When you click that button, the cursor will be between the tags, and you can enter your code.

 

[PeterOwen]

Wow, your son studies that?