Solve The Equation With Square Roots

Let $\displaystyle x = \sqrt{N-5},$ then we can rewrite the equation as:

$\displaystyle \sqrt{x^2 - 10x + 25} + \sqrt{x^2 + 14x + 49} = 12$

$\displaystyle \sqrt{(x-5)^2} + \sqrt{(x+7)^2} = 12$

$\displaystyle |x-5| + |x+7| = 12$

$\text{Case 1:}$

For $\displaystyle 0 \leq x \leq 5,$ imply $\displaystyle 5 \leq N \leq 30.$ We have the equation become $\displaystyle 5 - x + x + 7 =12.$ Hence we have $\displaystyle 30-5+1 = 26$ integer solutions.

$\text{Case 2:}$

For $\displaystyle x > 5,$ we have the equation become $\displaystyle 2x + 2 = 12$ imply $\displaystyle x = 5.$ Hence no solution.

Therefore from both cases, we have the answer is $\displaystyle \boxed{26}.$

Write the equation equivalently as $\sqrt{\left(\sqrt{N - 5} - 5\right)^2} + \sqrt{\left(\sqrt{N - 5} + 7\right)^2} = 12$ i.e. $\left|\sqrt{N - 5} - 5\right| + \left|\sqrt{N - 5} + 7\right| = 12$ Note that $N \geq 5$ . Consider three cases:

1) We have $\sqrt{N - 5} \geq 5 \Leftrightarrow N \geq 30$ , then $\sqrt{N - 5} - 5 + \sqrt{N - 5} + 7 = 12 \Leftrightarrow N = 30$ .

2) We have $-7 < \sqrt{N - 5} < 5 \Leftrightarrow N < 30$ , then $-\sqrt{N - 5} + 5 + \sqrt{N - 5} + 7 = 12 \Leftrightarrow 0 = 0$ , so $5 \leq N < 30$ .

3) We have $\sqrt{N - 5} \leq -7$ , then clearly $N \in \emptyset$ .

Taking the union of these three cases we find that $5 \leq N \leq 30$ , hence there are $\boxed{26}$ integer values of $N$ .

$\sqrt{N+20-10\sqrt{N-5}}+\sqrt{N+44+14\sqrt{N-5}}=12\\ ⇔ \sqrt{N-5-2*5*\sqrt{N-5}+25}+\sqrt{N-5+2*7*\sqrt{N-5}+49}=12\\ ⇔ \sqrt{(\sqrt{N-5}-5)^2}+\sqrt{(\sqrt{N-5}+7)^2}=12\\ ⇔ |\sqrt{N-5}-5|+|\sqrt{N-5}+7|=12\\ Apply\ the\ inequality\ |A|+|B| \geq |A+B| (equality\ occurs\ if\ AB\geq0),\ we\ can\ see\ that\ :\\ |\sqrt{N-5}-5|+|\sqrt{N-5}+7|=|5-\sqrt{N-5}|+|\sqrt{N-5}+7|\geq|5-\sqrt{N-5}+\sqrt{N-5}+7|=12.\\ As\ in\ the\ equation,\ equality\ occurs\,\ so\ (5-\sqrt{N-5})(\sqrt{N-5}+7)\geq 0 ⇔ -7\leq\sqrt{N-5}\leq5\\ 0\leq N-5\leq25 (\sqrt{N-5} is\ greater\ than\ or\ equal\ to\ 0\ with\ N\ being\ an\ integer)\\ ⇔ 5\leq N\leq30.\ There\ are\ (30-5)+1=26\ integer\ values\ of\ N\ that\ satisfy\ the\ equation.$ Sorry, I can't write a good solution. I haven't got used to writing it yet.

Part of the trick is that under each "big" radical, is the square of a simpler expression:

(√(N-5) - 5)² = (5 - √(N-5))² = N + 20 - 10√(N-5)

(√(N-5) + 7)² = N + 44 + 14√(N-5)

Another part of the trick is, that first squared expression can take either signed form. So you have two possible simplified forms:

√(N-5) - 5 + √(N-5) + 7 = 2√(N-5) + 2 = 12

√(N-5) = 5, N = 30, . . . or

5 - √(N-5) + √(N-5) + 7 == 12

. . an identity, and it applies as long as 5 - √(N-5) is ≥ 0. And for that, along with the stipulation that N be an integer, we must only have

0 ≤ N - 5 ≤ 25 N = {5, 6, .. , 30}

which are 26 integer values, and include the one we got on the first pass.

Take a look at the equation. Easily, we could simplified the equation into this state

$\sqrt{(\sqrt{N-5} - 5)^{2}} + \sqrt{(\sqrt{N-5} + 7)^{2}}$ = 12

$\sqrt{x^{2}} = |x|$ . Therefore, we have $|\sqrt{N-5}-5| + |\sqrt{N-5} + 7|$ . As $\sqrt{N-5} > 0$ , we could simplify the equation until we have

$|\sqrt{N-5} - 5| = 5 - \sqrt{N-5}$ which can only means $\sqrt{N-5} \leq 5$ .

With the last equation, or perhaps inequality, I must say, $5 \leq N \leq 30$ is the solution.