${\sqrt{2x+1}=1+\sqrt{x}}$ — I dont know if the solution is correct. Help?

Question

${\sqrt{2x+1}=1+\sqrt{x}}$
${2x+1=1+2\sqrt{x}+x}$
${x=2\sqrt{x}}$
${x*\frac{1}{x^{1/2}}=2}$
${\sqrt{x}=2}$
${x=4}$

Answer 1

You divided $x^{\frac12}$ without checking whether $x=0$ is a solution. When using division to solve an equation, you get only solutions where the divisor isn't zero, so be sure to separately check cases where the divisor is zero.

Answer 2

The fourth step is not correct. You may rather write $$ x=2\sqrt{x} $$ $$ x-2\sqrt{x}=0 $$$$ \sqrt{x}\left(\sqrt{x}-2\right)=0 $$ $$ \sqrt{x}=0 \quad \text{or} \quad \sqrt{x}-2=0 $$ giving easily $$x=0, \qquad x=4 $$ as solutions.

Answer 3

it must be $$x\geq 0$$ after squaring we get $$x=2\sqrt{x}$$ squaring again we obtain $$x(x-4)=0$$ thus we get $$x=0$$ or $$x=4$$ which are indeed solutions.

Answer 4

When you divide by $\sqrt{x}$, you are assuming something. This assumption discards another valid solution.

Answer 5

You are mostly correct but there are two conditions you didn't take into account which you should have.

$\sqrt{2x + 1}=1+\sqrt{x}$ Note, this implies $2x + 1 \ge 0$ i.e $x \ge -\frac 12$.

$2x +1 = 1 + 2\sqrt{x} + x$ Now, we have "lost" that assumption. It is possible that we will end up with some extraneous answers where $x < -\frac 12$. As it turns out that isn't an issue and it doesn't happen but it could have. (As we still have $\sqrt{x}$ that implies $x \ge 0$ so $x < -\frac 12$ is impossible).

$x = 2\sqrt{x}$

$x/x^{1/2} = 2$ Here you divided by $x^{1/2}$ in the assumption $x^{1/2} \ne 0$. You can not make that assumption. You must consider the possibility that $x^{1/2}$.

So say: Case 1: If $\sqrt{x} = 0$, then $x= 0$ and we have $0 = 2\sqrt{0}$ which consistent so $x = 0$ is a possible answer.

But if $\sqrt{x} \ne 0$ then

$x/x^{1/2} = 2$

$x^{1/2} = 2$

$x =4$

So $x=4$ is the only other possible solution . So $x = 0$ or $x = 4$.

Answer 6

For more information try WolframAlpha.

As for the solution:

$x=2\sqrt{x}$

$x-2\sqrt{x}=0$

$\sqrt{x}\left(\sqrt{x}-2\right)=0$

$\sqrt{x}=0 \quad \text{or} \quad \sqrt{x}-2=0$

We get:

$x=0, \qquad x=4$ as solutions.

Understood: you made the mistake in the 4th line

Answer 7

In addition to the omission of the solution $0,$ there's an important point here that I think some of the other answers are glossing over. Your method does not show that $0$ and $4$ are solutions — what it shows is that no number other than $0$ or $4$ is a solution. You still need to check that $0$ and $4$ actually work. (In this case, they do, but in some similar problems, you can get spurious solutions in addition to the actual solutions.)

Here's a complete argument:

If $${\sqrt{2x+1}=1+\sqrt{x}},$$ then deduce $$x=2\sqrt{x}$$ just as you did. Square both sides to get $$x^2=4x.$$ It follows that either (a) $x=0$ or (b) you can divide both sides by $x$ to get $x=4.$

At this point, all we have shown is that if $x$ is a solution, then $x$ must equal either $0$ or $4.$ So $0$ and $4$ are the only two possibilities for a solution. You still need to check out each one of these possible solutions individually:

$${\sqrt{2\cdot 0+1}=1=1+\sqrt{0}}$$

and

$${\sqrt{2\cdot 4+1}=3=1+\sqrt{4}},$$

so in fact both $0$ and $4$ are solutions (and so they're the only solutions, since we already know that any solution has to be either $0$ or $4).$