Solving radical equations x√3+x√2=1

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In summary: No, your son is not studying that. That is called The Freshman's Dream and it is a common error made by students.
  • #1
nek9876
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I'm trying to help my son with homework and can't get the following problem

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

Please help
 
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  • #2
nek9876 said:
I'm trying to help my son with homework and can't get the following problem

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

Please help
Start by factoring the x from the left hand side:
\(\displaystyle x \sqrt{3} + x \sqrt{2} = x \left ( \sqrt{3} + \sqrt{2} \right ) = 1\)

Can you finish from here?

-Dan
 
  • #3
nek9876 said:
I'm trying to help my son with homework and can't get the following problem

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

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. :)
 
  • #4
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}?
 
  • #5
nek9876 said:
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:

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

What is true though is:

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

Let's go back to:

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

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?
 
  • #6
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
 
  • #7
nek9876 said:
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

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

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:

\(\displaystyle x=\frac{1}{\sqrt{2}+\sqrt{3}}=\frac{1}{\sqrt{3}+\sqrt{2}}\cdot\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{3-2}=\sqrt{3}-\sqrt{2}\)

nek9876 said:
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 [MATH][/MATH] tags. When you click that button, the cursor will be between the tags, and you can enter your code.
 
  • #8
Wow, your son studies that?
 

FAQ: Solving radical equations x√3+x√2=1

What is a radical equation?

A radical equation is an equation that contains a radical, which is a mathematical symbol used to indicate the root of a number. The most common radicals are the square root (√) and the cube root (3√). Solving radical equations involves finding the value of the variable that makes the equation true.

What is the general process for solving radical equations?

The general process for solving radical equations involves isolating the radical on one side of the equation and then raising both sides to the appropriate power to eliminate the radical. This process is repeated until the variable is isolated and its value can be determined.

Why is it important to check for extraneous solutions when solving radical equations?

Extraneous solutions are solutions that do not satisfy the original equation, even though they may satisfy the simplified equation. This can occur when raising both sides of the equation to a power, as it can introduce solutions that were not present in the original equation. Checking for extraneous solutions ensures that the final solution is valid.

What are some common mistakes to avoid when solving radical equations?

One common mistake is forgetting to check for extraneous solutions. Another mistake is incorrectly simplifying the radical, which can lead to incorrect solutions. It is also important to be careful when raising both sides of the equation to a power, as this can introduce extraneous solutions.

How can I check my solution for a radical equation?

You can check your solution by plugging it back into the original equation and simplifying. If the resulting equation is true, then the solution is correct. It is also a good idea to double-check for any extraneous solutions.

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