Question

How would you solve this question or similar questions? It is solving for x.

Answer 1

Given the initial expression

`\sqrt[]{2x+3}=\sqrt[]{2x}+3`

Then,

`\begin{gathered} \sqrt[]{2x+3}=\sqrt[]{2x}+3 \\ \Rightarrow(\sqrt[]{2x+3})^2=(\sqrt[]{2x}+3)^2 \\ \Rightarrow2x+3=2x+6\sqrt[]{2x}+9 \end{gathered}`

Therefore,

`\begin{gathered} \Rightarrow3=6\sqrt[]{2x}+9 \\ \Rightarrow-6=6\sqrt[]{2x} \\ \Rightarrow\sqrt[]{2x}=-(6)/(6)=-1 \\ \Rightarrow\sqrt[]{2x}=-1 \\ \Rightarrow\sqrt[]{2}\sqrt[]{x}=-1 \\ \Rightarrow\sqrt[]{x}=-\frac{1}{\sqrt[]{2}} \end{gathered}`

And sqrt(x)>=0 for any real number.

Therefore, there is no real solution to the equation.