Helena solved the equation $$ \sqrt{x+7} =\sqrt{2x}+1,~x \in \mathbb{R} $$ in the following way.
(1) She squared both sides of the equation to get rid of the radical: $$ \begin{aligned} (\sqrt{x+7})^2 &=(\sqrt{2x}+1)^2 \cr x+7&=2x+2\sqrt{2x}+1\cr -x+6&= 2\sqrt{2x} \end{aligned}$$
(2) After simplifying, she obtained an equation that still has a square root, so she squared both sides of the equation again: $$ \begin{aligned} (-x+6)^2&=(2\sqrt{2x})^2 \cr x^2-12x+36&=8x \cr x^2-20x+36&=0 \end{aligned}$$
(3) Finally, Helena got a quadratic equation, which she solved by quadratic formula: $$ x_{1,2}=\frac{20\pm \sqrt{(-20)^2-4 \cdot 1 \cdot 36}}{2}=\frac{20\pm \sqrt{256}}{2}=\frac{20\pm 16}{2} $$ The solutions of the given radical equation are: $$ x_1=18, x_2=2 $$ Do you find errors in her solution or is her solution perfectly fine?
Helena did not proceed correctly. She forgot to check each solution in the original equation. The check is necessary because the solutions of the new equation obtained by squaring may not be valid solutions of the original equation.
Helena's solution is perfectly fine.
Helena made errors in step (1) and (2). It is not allowed to square both sides of the equation.
She made an error in step (1) when squaring $\sqrt{2x}+1$. It should have been: $(\sqrt{2x}+1)^2=2x+1$.
The quadratic formula used in step (3) is wrong. The right formula is: $$ x_{1,2}=\frac{-20\pm \sqrt{(−20)^2−4\cdot 1 \cdot 36}}{2} $$
The check for $x=18$: $$ L=\sqrt{18+7}=\sqrt{25}=5,~R=\sqrt{2 \cdot 18}+1=\sqrt{36}+1=6+1=7⇒L\neq R $$ The check for $x=2$: $$ L=\sqrt{2+7}=\sqrt{9}=3,~R=\sqrt{2\cdot 2}+1=\sqrt{4}+1=2+1=3⇒L=R $$ The check shows that the original equation has only one solution, $x=2$.