Wikipedia:Reference desk/Archives/Mathematics/2017 July 6

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July 6

Solve for x

On a website ad I saw the equation:

${\sqrt {x+15}}+{\sqrt {x}}=15$

Using a spreadsheet I found out the solution, but how would I solve this using algebra?--Commander Keane (talk) 06:35, 6 July 2017 (UTC)

One way is to substitute $x=y^{2}$ into the equation and solve for y. Another way is to square both sides of the equation, isolate the remaining square root term on one side, square again, then solve the resulting quadratic.--Wikimedes (talk) 07:24, 6 July 2017 (UTC)

Thank you so much for your reply :-) Both methods work, but in doing the second one I did not need to solve a quadratic, unless I missed something.--Commander Keane (talk) 08:40, 6 July 2017 (UTC)

Yes, you were saved when the x² terms canceled out. A third variation of the method is to move the ${\sqrt {x}}$ term to the right and square both sides. Now the x terms cancel out and you have a simple equation for ${\sqrt {x}}$.

It is also possible to find a solution mentally without using algebra, by seeing whether ${\sqrt {x}}$ might be an integer. If it is, then x and x+15 are two square numbers that differ in size by 15. The differences between successive squares are the successive odd numbers, so you can easily see that the two squares must be either 1 and 16 (which are 3+5+7=15 apart) or 49 and 64 (which are 15 apart). If any larger squares were involved they would be more than 15 apart. x=1 does not work, but x=49 does. This approach finds a solution but does not prove that it is the only solution, though. --76.71.5.114 (talk) 09:10, 6 July 2017 (UTC)

The uniqueness follows from that fact that ${\sqrt {x+15}}+{\sqrt {x}}$ is an increasing function of $x$. Another solution that I saw somewhere (probably Quora) is to define ${\sqrt {x+15}}-{\sqrt {x}}=y$. Multiplying the two equations shows that $y=1$, and then subtracting gives $2{\sqrt {x}}=14$, and hence $x=49$. AndrewWTaylor (talk) 10:09, 6 July 2017 (UTC)