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

Question

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

The statement tells us that $x\in[5,9]$. Also,

$$\sqrt{x-5}\gt1+\sqrt{9-x}$$

Since both sides are positive, we can square

$$x-5>1+9-x+2\sqrt{9-x}\\2x-15\gt2\sqrt{9-x}$$

$\implies 2x-15\gt0\implies x\gt7.5$

Since $x\in\mathbb Z\implies x=8,9$

But on back substitution, $x=8$ doesn't satisfy. Is there a way we could get the final answer without back sustitution?

Answer 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 $x = 9$.

Answer 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 $(x-a)(x-b)<0, a<b;$ $$a<x<b$$