$$\sqrt{3x+1} - \sqrt{6-x} +3x^2-14x-8=0 $$
I tried : $3x+1 = a $ , $6-x=b $ and tried to make $ 3x^2-14x-8 $ to be in term of $a$ and $b$, I'm unable to solve it so far.
but I know the answer is $ x = 5 $ by trying to guess because $ 6-x \geq 0 $ and I checked the integer from $ 0$ to $6$.
I can't think of a way to solve this properly without guessing the answer.
On
I've done some reverse-factoring ( don't know if this is the right word or not ) knowing the answer is 5
so i got
$(x-5)\left [(\frac{\sqrt{3x+1}-4}{(x-5)})-(\frac{\sqrt{6-x}-1}{(x-5)})+(\frac{(3x+1)(x-5)}{(x-5)}) \right ] = 0 $
$(x-5)\left [(\frac{3}{(\sqrt{3x+1}+4)})+(\frac{1}{(\sqrt{6-x}+1)})+(3x+1) \right ] = 0 $ the right bracket is always more than $0$
i guess this is one way to do it lol.
The function $f(x)=3x^2-14x-8$ is strictly increasing, for $x\geq -7/3$ and the function $g(x)=\sqrt{3x+1}-\sqrt{6-x}$ is defined for $x\in[-{1\over 3},6]$ and is also increasing.
So equation $f(x)+g(x)=0$ has at most one solution. Can you guess it?
Edit: $6-x$ is decreasing, so is $\sqrt{6-x}$ but then $-\sqrt{6-x}$ is increasing. Since $\sqrt{3x+1}$ is increasing and the sum of incrasing function is increasing, $g(x)$ must be increasing.