sqrt(x + 23) + sqrt(x - 26) = 7

Question

anonymous · Answer

gotta square twice. ick

kinggeorge · Answer

So you want to find some x such that $x+23$ is a perfect square, and $x-26$ is a perfect square. In other words, you want a perfect square $n$ such that $n+23+26=n+49$ is also a perfect square. 

Fortunately, we can notice that $n=0$ fulfills our requirements very well. Since 0 is a perfect square, and $0+49=49=7^2$. Thus, we want $x-26=0$. Solving this, we get $x=26$

anonymous · Answer

is it because x 26 would be positive

anonymous · Answer

i am not understanding how the answer is 26

kinggeorge · Answer

Well, $x-26$ must be a perfect square, and $x+23$ must be a perfect square. Now let $n=x-26$. 

This means that $x+23=n+26+23=n+49$. 

Now we know that $n$ is a perfect square, and $n+49$ is a perfect square. From a quick glance, we see that if $n=0$, both properties hold. 

From there, we backtrack a little bit to where we know $n=x-26$. Since $n=0$, we have that $x-26=0$ so $x=26$

anonymous · Answer

how do we do this problem: Solve.
sqrt(5 x) - sqrt(5 x - 21) = 3

kinggeorge · Answer

I need to go now. Just post the problem in a new thread, and someone should be able to help you.