Help with minor maths puzzle

This isn’t homework (I just got my pensioner’s bus pass :cool:), but comes from a newspaper puzzle.

There was an equation with square roots on both sides and you had to simplify/solve it.
It was something like:

square root of (a+b) + square root of (a) = square root of (a+21),

where a and b are integers.

OK, by inspection a=4 and b=5.
But how would I square both sides of the equation?

(After squaring, I get a+b + a + 2(square root of (a+b) + square root of (a)) = a+21.

This is using the algebra (a+b) squared = a squared + b squared + 2ab.

(Why do I need to click the Mult-quote + these days instead of just Quote? Do others have this symptom, or is it just one of my viruses at work?)

I think the “+” I underlined should be a multiplication.

To get rid of the remaining square-root, just move everything else to the other side of the equation and square each side again. (Here, however, that just leaves you with something that seems over-complicated.)

Your puzzle has two unknowns, so there may be multiple answers. I’ve recently seen this puzzle here and there “square root of (x+15) + square root of (x) = 15” in which the trick is to move the square root of x to the other side of the equals before squaring Might that one be the one you saw…

I used Wolfram Alpha:

b = a - 2sqrt(a + 21)sqrt(a) + 21

[I don’t have Wolfram Alpha Pro, so I can’t see the steps it used to transform your equation into the one above.]

So while a = 4, b = 5 is a solution, any pair of values (a, b) that satisfy the above equation are also solutions.

For example, let a = 10.

Then b = 10 - 2sqrt(31)sqrt(10) + 21 = 31 - 2sqrt(310), so (10, 31 - 2sqrt(310)) is also a solution.

(A bit harder to see by inspection…)

Making the correction septimus indicated, combining the two multiplicands to get one square root, isolating that root and squaring it is the solution method but does leave you with something more complicated, arguably, then what you started with.

The essential point of the puzzle is that a and b are integers. I assume that a and a+b must be non-negative. Note that a(a+21) is a perfect square. So all possible solutions (a,b) are (0,21), (4,5), (7,0), (27,-24), and (100,-99).

Indeed - I overlooked that part. Thanks.

Thanks to all who replied!

Andy is right - I muddled up the original puzzle. :o

I suppose this wasn’t the puzzle you intended, it turns out, but for what it’s worth:

In other words, sqrt(a + b) = sqrt(a + 21) - sqrt(a). In other words, b = (sqrt(a + 21) - sqrt(a))^2 - a. So any choice of “a” gives rise to a solution, picking the appropriate value of “b”, if we’re to allow ourselves arbitrary values without whole number restrictions.

I was going to suggest that the OP puzzle, slightly modified so that a and b are required to be positive integers, does make a nice puzzle since there is indeed a unique solution. More fun than the “official” puzzle.

Glad to help.

Square the original expression and subtract a from each side:
a+b + sqrt(a(a+b)) = 21
That implies a+b can only be an integer from 1 to 21. Some more manipulation gives:
4a = (a+b) - 42 + 441/(a+b)
Thus (a+b) must be a factor of 441=3377. There are five possible factors meeting the requirement of being 21 or less: 1, 3, 7, 33, and 3*7
Inserting them into the expression for 4a results in the 5 solutions given by DPRK.

That’s (probably) all on you & your set-up. I (and probably the rest of us) don’t have that symptom.