Find the roots of the equation,
$$\frac{(7-x)\sqrt{7-x} + (x-5)\sqrt{x-5}}{\sqrt{7-x} +\sqrt{x-5}}=2$$ What I did till now $(7-x)^{3/2}+(x-5)^{3/2}/(7-x)^{1/2}+(x-5)^{1/2}$ Now what to do next? Please help me out. I am stuck at this question since an hour
On
Hint:
Notice that $$A^3+B^3=(A+B)(A^2-AB+B^2)\qquad\text{for real numbers }A\text{ and }B$$ So $$\frac{A^3+B^3}{A+B}=A^2-AB+B^2$$
On
Let $u=\sqrt{7-x},\;v=\sqrt{x-5}$. \begin{align*} \text{Then}\;\;&\frac{(7-x)\sqrt{7-x}+(x-5)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}=2\\[4pt] \implies\;&\frac{u^3+v^3}{u+v}=2\\[4pt] \implies\;&\frac{(u+v)(u^2-uv+v^2)}{u+v}=2\\[4pt] \implies\;&u^2-uv+v^2=2\\[4pt] \implies\;&uv=u^2+v^2-2\\[4pt] \implies\;&uv=(7-x)+(x-5)-2\\[4pt] \implies\;&uv=0\\[4pt] \implies\;&u=0\;\;\text{or}\;\;v=0\\[4pt] \implies\;&x=7\;\;\text{or}\;\;x=5\\[4pt] \end{align*} So the only possible solutions are $x=7$ or $x=5$.
If you substitute into the original equation, you can determine which of those work (if any).
In fact, they both work.
On
Directly solving the equation,
$$\begin{align*} \frac{(7-x)\sqrt{7-x}+(x-5)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}} &= 2\\ (7-x)\sqrt{7-x}+(x-5)\sqrt{x-5} &= 2\left(\sqrt{7-x}+\sqrt{x-5}\right)\\ (5-x)\sqrt{7-x}+(x-7)\sqrt{x-5} &= 0\\ \sqrt{x-5}\sqrt{7-x}(\sqrt{x-5}+\sqrt{7-x}) &= 0 \end{align*}$$
The first two factors imply $x=5$ or $x=7$, and the third factor is always non-zero.
Hint: $A^3+B^3=(A+B)(A^2-AB+B^2)$