B2. Find all real solutions of the equation

$$
\sqrt{x+4}+\sqrt{2 x+1}+\sqrt{4 x+2}=\sqrt{3-x}
$$

## 61st Mathematical Competition for High School Students in Slovenia

Selection Competition, March 16, 2017

## Problems for 3rd Year

## Time for solving: 45 minutes.
B2. We transform the equation

$$
\sqrt{2 x+1}+\sqrt{x+4}=\sqrt{3-x}-\sqrt{4 x+2}
$$

by squaring it

$$
(2 x+1)+2 \sqrt{2 x+1} \sqrt{x+4}+(x+4)=(3-x)-2 \sqrt{3-x} \sqrt{4 x+2}+(4 x+2)
$$

and simplifying, we get

$$
\sqrt{2 x+1} \sqrt{x+4}=-\sqrt{3-x} \sqrt{4 x+2}
$$

When we square the equation again and simplify, we get the quadratic equation

$$
6 x^{2}-x-2=0
$$

Using the quadratic formula, we find the solutions \( x_{1}=-\frac{1}{2} \) and \( x_{2}=\frac{2}{3} \). We test both solutions. The first solution satisfies the equation, but the second does not, as we get \( 2 \sqrt{\frac{14}{3}} + \sqrt{\frac{7}{3}} = \sqrt{\frac{7}{3}} \). Therefore, the only solution to the equation is \( x = -\frac{1}{2} \).

Writing the equation \(\sqrt{2 x+1}+\sqrt{x+4}=\sqrt{3-x}-\sqrt{4 x+2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .1\) point

Correct squaring of the equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 points

![](https://cdn.mathpix.com/cropped/2024_06_07_9b375d7a0e2f6834f525g-14.jpg?height=60&width=1579&top_left_y=1329&top_left_x=273)

Squaring and rearranging the equation to \( 6 x^{2}-x-2=0 \). ..................... 2 points

Factoring the quadratic equation. ...........................................................................................................

Writing both solutions \( x_{1}=-\frac{1}{2} \) and \( x_{2}=\frac{2}{3} \). .................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_9b375d7a0e2f6834f525g-14.jpg?height=60&width=1579&top_left_y=1535&top_left_x=273)

![](https://cdn.mathpix.com/cropped/2024_06_07_9b375d7a0e2f6834f525g-14.jpg?height=66&width=1579&top_left_y=1583&top_left_x=273)

2. method. In the same way as in the first solution, we get

$$
\sqrt{2 x+1} \sqrt{x+4}=-\sqrt{3-x} \sqrt{4 x+2}
$$

We see that the left side of the equation is greater than or equal to 0, and the right side is less than or equal to 0. Therefore, both sides must be equal to 0, which is only possible for \( x = -\frac{1}{2} \).
Writing the equation \(\sqrt{2 x+1}+\sqrt{x+4}=\sqrt{3-x}-\sqrt{4 x+2}\) ..... 1 point
Correct squaring of the equation ..... 4 points
Writing the equation \(\sqrt{2 x+1} \sqrt{x+4}=-\sqrt{3-x} \sqrt{4 x+2}\). ..... 5 points
Observing that the left side of the equation is greater than or equal to 0,
and the right side is less than or equal to 0. ..... 2 points
Conclusion: \((2 x+1)(x+4)=0\) and \((3-x)(4 x+2)=0\) ..... 2 points
Writing the solutions \( x_{1}=-\frac{1}{2} \) and \( x_{2}=-4 \). ..... 1 point
Writing the solutions \( x_{3}=-\frac{1}{2} \) and \( x_{4}=3 \). ..... 1 point
Excluding the solutions \( x=-4 \) and \( x=3 \) ..... 2 points
Writing the solution \( x=-\frac{1}{2} \) ..... 2 points

3. method. We observe that \( 4 x+2=2(2 x+1) \), so we can transform the equation into

$$
(1+\sqrt{2}) \sqrt{2 x+1}=\sqrt{3-x}-\sqrt{x+4}
$$

We square the equation and rearrange it to get

$$
(3+2 \sqrt{2}) x+\sqrt{2}-2=-\sqrt{3-x} \sqrt{x+4}
$$

We square it again and rearrange to get

$$
(18+12 \sqrt{2}) x^{2}+(-3-2 \sqrt{2}) x-6-4 \sqrt{2}=0
$$

We observe that the equation can be simplified by \( 3+2 \sqrt{2} \), to get

$$
6 x^{2}-x-2=0
$$

As in the first solution, we find that this equation has solutions \( -\frac{1}{2} \) and \( \frac{2}{3} \), but only the first solution satisfies