simplify this expression

[Prasad]

[math]\sqrt{8+2\sqrt{7}}[/math]+[math]\sqrt{8-2\sqrt{7}}[/math]

[mooeypoo]

Moved to homework help.

 

Prasad, how would you start this equation? We're not going to solve it for you, but we can help you solve it yourself. Do you have any idea on how to start?

 

Try to look for common numbers in these groups.

[dttom]

Square root is troublesome, now if you don't want root what would you do?

[Vindhya]

I am a pathologist 's daughter trying to help my friend. please help me to find the answer.

[mooeypoo]

Vindhya, we would love to help, but we don't want to just give you the answer (that's not help).

 

Here's a tip, see if that helps your kid continue:

 

You start with this expression:

 

[math]\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}[/math]

 

Ignore, for a moment, the "scary" square root that covers both main parts. What if this looked like this:

[math](8+2\sqrt{7})[/math]

and

[math](8-2\sqrt{7})[/math]

 

The first thing that pops to me, as a start, is that both expressions have a 2 multiplying the second part (the square root inside). It would be nice if I could find a way to multiply the first part of each expression by 2, so I can have a common multiplier I can then take outside...

 

Like this:

[math](2(4)-2\sqrt{7})[/math]

 

Don't be afraid of square roots, they just group the elements together. Now that you see how to stat, you can bring back the square root again:

[math]\sqrt{2(4)-2\sqrt{7}}[/math]

 

Can you do the same for the other part and try to continue from here?

[D H]

You want to simplify

 

[math]x=\sqrt{8+2\sqrt 7} + \sqrt{8-2\sqrt 7}[/math]

 

What is x²?

[mooeypoo]

You want to simplify

 

[math]x=\sqrt{8+2\sqrt 7} + \sqrt{8-2\sqrt 7}[/math]

 

What is x²?

I don't mean to confuse here, but I was always under the impression that tricks like these are problematic. When you square the x, solve and then find the sqrt, you gain signs (+/-). It might be a problem for the "more advanced" but whenever we do something like that in physics or math, we end up with an absolute value. I was always told to avoid doing that.....?

[D H]

Absolutely not.

 

Squaring does not introduce any ambiguity because the square root symbol denotes the positive square root, always. There is only one answer to the question "what is x²?" Since the original question obviously has a positive answer, going from "what is x²?" to "what is x?" is similarly unambiguous.

[mooeypoo]

Absolutely not.

 

Squaring does not introduce any ambiguity because the square root symbol denotes the positive square root, always. There is only one answer to the question "what is x²?" Since the original question obviously has a positive answer, going from "what is x²?" to "what is x?" is similarly unambiguous.

Right, also we deal with numbers here and not variables.. hm.

 

Okay, so the ambiguous part would be asking to square something like

[math]x=\sqrt{y+3}[/math] that has variables? I'm a bit confused, now. When do we gain/lose signs and should be careful of squaring the roots?

[Cap'n Refsmmat]

Whenever you may be interested in the negative root as well as the positive root.

[mooeypoo]

I'm going to look up the examples, I can't remember them now, but there were a few where squaring the equation was the wrong thing to do, because it yielded two solutions rather than one, which meant we need to verify which is the right one later.

[timo]

You do that here, too. You can solve for x^2 and then consider which of the two candidate x is the correct one. D_H just put that on the start of the statement "since the original question obviously has a positive answer, going from 'what is x^2?' to 'what is x?' is similarly unambiguous". If the equation was [math]x = -\sqrt{\dots}[/math] then you'd get the same x^2 but chose the negative candidate for x "since the original question obviously has a negative answer".

Of course you should also pay attention to the range that the original term is defined over, but that problem is not unique for squaring but also appears in other operations.

[swansont]

I'm going to look up the examples, I can't remember them now, but there were a few where squaring the equation was the wrong thing to do, because it yielded two solutions rather than one, which meant we need to verify which is the right one later.

 

They probably involve variables, where you don't know the sign, rather than just numbers.

[D H]

Anyhoo, back to the problem at hand:

 

Prasad and Vindhya, if you are still around, I strongly suggest you try the hint in post #6.

---

Merged post follows:

Consecutive posts merged

And now back to mooeypoo's concern:

 

Any multivalued function (e.g., [math]x^{1/2}, \arcsin(x), \cdots[/math]) can lead to potential problems. That doesn't mean one should steer clear of them. You just need to pay a bit more attention.

 

For example, suppose you are asked to find the sum of the two positive solutions to

 

[math]x^4 - 16 x^2 + 36 = 0[/math]

 

That there are negative solutions is not all that much of an issue. You just need to pay a bit more attention than you do with simpler problems.

Edited March 15, 2010 by D H
Consecutive posts merged.

[mooeypoo]

For example, suppose you are asked to find the sum of the two positive solutions to

 

[math]x^4 - 16 x^2 + 36 = 0[/math]

 

That there are negative solutions is not all that much of an issue. You just need to pay a bit more attention than you do with simpler problems.

Right!

 

But let's say you create a model of something (like, say, model some physical movement using forces, lagrangians, whatever), and you got something with variables and even roots. You don't know, necessarily, which answer (positive or negative) to expect, you just know you want to account for the values of x. If you square the result, you *ADD* values to x, that wouldn't have been there previously. So at the end, you will have to verify each result if it can stand the original equation or not.

 

Err, I hope I am getting the point across right. I agree (and thx DH for the explanation you sent) that in this particular question, because we deal with finite numbers, the concern is moot, but this is something that I thought held, at least for variables.

 

Anyhoo if we continue debating this, we could split it off to its own subject. I was just a bit confused, I remember both my physics and math professors warning against squaring results and when we do, we need to take into account absolute-value answers.

 

It's a bit higher level than the current question, I guess.

 

~moo

[D H]

For example' date=' suppose you are asked to find the sum of the two positive solutions to

 

[math']x^4 - 16 x^2 + 36 = 0[/math]

 

That there are negative solutions is not all that much of an issue. You just need to pay a bit more attention than you do with simpler problems.

 

Right!

The above question is very relevant to the discussion at hand. The four solutions are given by

 

[math]x = \pm\sqrt{8 \pm 2\sqrt 7}[/math]

 

The two positive solutions sum to [math]\sqrt{8 + 2\sqrt 7} + \sqrt{8 - 2\sqrt 7}[/math] -- which is the problem raised in the original post.

[Vindhya]

Thank you for your help. This was a 2-mark problem for 10th standard student. I just split the expressions and the final answer I got was 2\sqrt 7. Please tell me whether I am correct in my approach.

[Cap'n Refsmmat]

Yeah, I believe that's right.

[D H]

That is correct.