Solve $\sqrt{x-5}-\sqrt{9-x}\gt1,x\in\mathbb Z$

Solution 1:

Starting from where you left off, since both sides of the inequality $$2x - 15 > 2\sqrt{9 - x}$$ are positive, the direction of the inequality is preserved if we square both sides, which yields \begin{align*} 4x^2 - 60x + 225 & > 4(9 - x)\\ 4x^2 - 60x + 225 & > 36 - 4x\\ 4x^2 - 56x & > -189 \end{align*} Since $(2a + b)^2 = 4a^2 + 4ab + b^2$, we can complete the square with $a = x$ and $b = 14$ to obtain \begin{align*} 4x^2 - 56x + 196 & > 7\\ 4(x^2 - 14x + 49) & > 7\\ (x - 7)^2 & > \frac{7}{4} \end{align*} Since $(8 - 7)^2 = 1 < \dfrac{7}{4}$, this eliminates $8$. Thus, the only integer solution is $x = 9$.

Solution 2:

Hint:

Let $\sqrt{x-5}=a\ge0$ and $\sqrt{9-x}=b\ge0$

$$\implies a^2+b^2=4$$

and $a-b>1\iff a> b+1$

$$4=a^2+b^2>(b+1)^2+b^2\iff 2b^2+2b-3<0$$

Now for $(x-a)(x-b)<0, a<b;$ $$a<x<b$$