Find the Value of X - what tool is best?

My wife likes to solve puzzles and found this one a few days ago.

Find X in this equation:

Sqrt(x+15)+Sqrt(x)=15

She worked a bit on the problem and failed. She brought it to me as I once taught math in high school.

The first problem is to find X, but the bigger question is what method did you use to solve it?

If you are up to the challenge, could you find X and then tell me how you solved it!!

I'll give you my method after I get a few replies.

We have a first reply from mike7952 who uses the webpage of Tiger-Algerbra.com .

The answer pops out by simply typing in the question. See it at:

http://www.tiger-algebra.com/drill/S...5)_Sqrt(x)=15/

Method 1 = find the web page that does the work for us. Not bad - rep would be given if he replied to this post, but he PMed me so no rep given.

Well, didn't know I was allowed to cheat lol

Hey Mike,

Rep given as this is one of many ways to get an answer. I think with our modern technology, asking Alexa, Siri or Cortana might be the preferred method in the future.

Looks like tim (above) simply put the question on a puzzle page. This is a second method and rep would be given if it produces an answer. Asking the question a second time doesn't produce a rep. It is interesting to me that the first two answers are internet based. I wonder when someone will pull out a pencil and paper to do this?

Are the first two methods really - ask someone else or something else?

No rep for cytop as s/he didn't give an answer to the value of x or even a method. Sorry cytop. Finding a method is what I'm after in this post.

I could of got better grades in school if only I had the tools the kids have these days. .

I never give negative rep. (I think I did once by clicking on the wrong radio button). Still to the point - what method would you use?

I was thinking about a graphing program after moving the 15 to the left and setting it equal to y. Then see where the graph went and where it might cross the x axis.
I also was thinking about the method called "completing the square" which I taught in High School math.
I also wondered if changing the "Sqrt" to powers of 1/2 and see if the problem look friendlier. That didn't seem to help either.

cytop - do you have a preferred method to solve this problem (back to the topic)?

Originally Posted by MarvinP

If you are up to the challenge, could you find X and then tell me how you solved it!!

I'll give you my method after I get a few replies.

Result: 49

You have Inverse function calculator built in excel:

Look gif here:
https://online-excel-solutions.com/t...n-in-excel/185

Explanation:

1. Select cell with result or change it in Set cell
2. Go to Data -> What-if Analysis -> Goal Seek
3. Enter desired result in To Value field
4. Choose By changing cell to pick up input value that Excel can adjust
5. click OK

In your case (Formula is in B1, variable is in A1) it will look like this:

Untitled.png

Rep for Zbor, who actually used Excel and Solver to get an answer. Rep given. Thanks for another method!!

Thx for the rep

Math solution:

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Fun problem.

Zbor's solutions (using Goal Seek and algebra) are the two first solutions I would have come up with, so I won't exactly repeat his suggestions.

The one "problem" I had with Zbor's solutions is that he seemed to limit the solution to real numbers. I'm not too afraid of complex numbers, and I wanted to convince myself that there was not a second solution (as a lot of square root and quadratic problems have). So, I pulled out my Newton-Raphson tutorial (https://www.excelforum.com/tips-and-...ind-roots.html I actually used the secant method, because I did not trust myself to get the exact expression for dy/dx without error) and put this problem into the complex number variation. I tried several widely varying starting values, and all attempts converged on the 49+0i solution that Zbor found. I have not fully proven to myself that there is only one unique solution to this problem, but it seems that there is. I don't know if "programming your own numeric root finding algorithm rather than using the built in Goal Seek" is considered different enough from zbor's suggestion.

If we limit to real numbers, I could suggest a graphical solution -- plot several values for f(x) and see where it crosses f(x)=15. This can be a useful tool for determining if there might be multiple real solutions or not.

As I thought further about the problem, one aspect of the problem jumped out at me. The problem adds two irrational numbers (at least, most values for x would yield irrational numbers) to get an integer. I have no proof for this, but it seems very unlikely, if not impossible, for two irrational numbers to sum up to an integer. Suggesting that each SQRT() term should be the SQRT() of a perfect square. This suggests that, if one wanted to avoid the high math of these other solutions -- could you solve the problem by "intuitively" guessing which two perfect squares have square roots that add up to 15. 7+8=15 (the solution found here) is fairly easy to back out without a calculator.

Try 36 and 81 (6+9) and you can quickly see that 81 is much more than 15 greater than 36 -- which led to something like a "system of equations" solution. Find two perfect squares (a and b) where the difference between a and b is 15 (b=a+15) and where the square roots of those numbers add up to 15. Again, I don't know that there is a good proof that this is always true, but it does work for this problem.

OK - it seems this thread has lost interest, so here is how I solved this problem....

I guessed the correct answer, plane and simple, in about 1 minute. I looked at sqrt(x) and tried 25 then 36 then 49 and 64. I was surprised to see 36 + 15 = 64. No algebra, no trig, no Excel Solver. I was proud and ashamed of myself at the same time. What kind of method is "guessing". Then I remembered a class I took while teaching in about 1978 from Dr. Anderson who was a protégée of Charles Polya (one of the last great mathematicians). The class was called "How to Solve It" and one of the first methods was to guess. I guess I learned something from that class!!??

Read more about solving problems and teaching at:
https://www.scribd.com/document/1372...ng-Mathematics


BTW - my wife did come in with the algebra answer but she did it on her own, without looking at any web site.

I was surprised and happy to see I didn't need to spend more time on the problem, as I had guessed an answer.

I moved the radical x to the 15 side and the squared both sides
I got at the end -210=-30radx and 7=radx and x=49

Sorry marvin but I don't know how to post the math type

Hey Orrin,

I don't understand your answer. Or is it a question?

Look again, I uploaded a sheet with it.

Maybe VBA ?

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or array formula
=MATCH(0,(SQRT((ROW(INDIRECT("A1:A1000")))+15)+SQRT((ROW(INDIRECT("A1:A1000")))))-15,1)

@Pike

I'd use:

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@Marvin

no benefit, resulting from refinement:

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Thinking about this problem, and I discovered something interesting (I would not be surprised if I had proved this as a homework exercise years ago, but not remembered it -- maybe this is how your subconscious caught on to this so quickly).

More generic expression for problem: sqrt(x+a)+sqrt(x)=a assuming perfect squares and integers and such, I noticed but did not think much about how 15 was present twice in the problem. From the generic statement, I wondered if there are other values for a that would also work (again, assuming perfect squares and integers and such). Note that a different way of expressing the problem is "find two perfect squares [x and x+a] whose difference [a] is the sum of their square roots". Observations:
x=1 a=3 works (1 and 4 are the perfect squares 4-1=3=1+2)
x=2 a=5 works (4 and 9 are the squares 9-4=5=2+3)
and so on. One could put any odd number in place of the 15, and there will be two consecutive perfect squares that solve the problem.

A very interesting property of square numbers.

Originally Posted by MrShorty

Thinking about this problem, and I discovered something interesting (I would not be surprised if I had proved this as a homework exercise years ago, but not remembered it -- maybe this is how your subconscious caught on to this so quickly).

More generic expression for problem: sqrt(x+a)+sqrt(x)=a assuming perfect squares and integers and such, I noticed but did not think much about how 15 was present twice in the problem. From the generic statement, I wondered if there are other values for a that would also work (again, assuming perfect squares and integers and such). Note that a different way of expressing the problem is "find two perfect squares [x and x+a] whose difference [a] is the sum of their square roots". Observations:
x=1 a=3 works (1 and 4 are the perfect squares 4-1=3=1+2)
x=2 a=5 works (4 and 9 are the squares 9-4=5=2+3)
and so on. One could put any odd number in place of the 15, and there will be two consecutive perfect squares that solve the problem.

A very interesting property of square numbers.

This part of the Common Core Math?

I don't know if it is or not.

Originally Posted by MrShorty

Thinking about this problem, and I discovered something interesting (I would not be surprised if I had proved this as a homework exercise years ago, but not remembered it -- maybe this is how your subconscious caught on to this so quickly).

More generic expression for problem: sqrt(x+a)+sqrt(x)=a assuming perfect squares and integers and such, I noticed but did not think much about how 15 was present twice in the problem. From the generic statement, I wondered if there are other values for a that would also work (again, assuming perfect squares and integers and such). Note that a different way of expressing the problem is "find two perfect squares [x and x+a] whose difference [a] is the sum of their square roots". Observations:
x=1 a=3 works (1 and 4 are the perfect squares 4-1=3=1+2)
x=2 a=5 works (4 and 9 are the squares 9-4=5=2+3)
and so on. One could put any odd number in place of the 15, and there will be two consecutive perfect squares that solve the problem.

A very interesting property of square numbers.

If you put y instead of 15 and solve the equation you'll get that

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.

So if x = 49 then y = 15.
If x = 121 then y = 23
If x = -1 then y = 1+2*i
and so on...

@ snb
of course no need for indirect but if the worksheet has other calx's
=MATCH(15,SQRT(ROW(A1:A1000))+SQRT(ROW(A1:A1000)+15),0)
to reduce the volatile recalculation to column A
other array formula variations

I thought you more an index dude
=MATCH(15,SQRT(ROW(INDEX(A:A,1):INDEX(A:A,1000))+15)+SQRT(ROW(INDEX(A:A,1):INDEX(A:A,1000))),1)
or
=MATCH(15,SQRT(ROW(OFFSET(A1,,,1000))+15)+SQRT(ROW(OFFSET(A1,,,1000))),1)