given that (a\neq b), is it ever possible to have (sqrt{a}+sqrt{b}=sqrt{a + b})? explain.

Explanation:

Step1: Square both sides

Square the equation $\sqrt{a}+\sqrt{b}=\sqrt{a + b}$. We get $(\sqrt{a}+\sqrt{b})^2=(\sqrt{a + b})^2$. Using the formula $(x + y)^2=x^{2}+2xy + y^{2}$, the left - hand side is $a + 2\sqrt{ab}+b$, and the right - hand side is $a + b$. So, $a + 2\sqrt{ab}+b=a + b$.

Step2: Simplify the equation

Subtract $a$ and $b$ from both sides of the equation $a + 2\sqrt{ab}+b=a + b$. We obtain $2\sqrt{ab}=0$.

Step3: Solve for $a$ and $b$

Divide both sides of $2\sqrt{ab}=0$ by 2 to get $\sqrt{ab}=0$. Then, square both sides to find $ab = 0$. This means either $a = 0$ or $b = 0$. If $a = 0$ and $b\neq0$, the original equation becomes $\sqrt{0}+\sqrt{b}=\sqrt{0 + b}$, which simplifies to $\sqrt{b}=\sqrt{b}$. Similarly, if $b = 0$ and $a\neq0$, $\sqrt{a}+\sqrt{0}=\sqrt{a + 0}$, which simplifies to $\sqrt{a}=\sqrt{a}$.

Answer:

Yes, it is possible when either $a = 0$ and $b\neq0$ or $b = 0$ and $a\neq0$.