Solve the following equation: sqrt (3x+4)-sqrt (2x+1)=1 No real solutions 0 0,4 4

To solve the equation $\sqrt{3x+4} - \sqrt{2x+1} = 1$, we can start by isolating one of the square roots.<br /><br />First, let's isolate $\sqrt{3x+4}$:<br />$\sqrt{3x+4} = \sqrt{2x+1} + 1$<br /><br />Next, square both sides of the equation to eliminate the square root:<br />$(\sqrt{3x+4})^2 = (\sqrt{2x+1} + 1)^2$<br />$3x+4 = (2x+1) + 2\sqrt{2x+1} + 1$<br />$3x+4 = 2x+2 + 2\sqrt{2x+1}$<br />$x+2 = 2\sqrt{2x+1}$<br />$(x+2)^2 = 4(2x+1)$<br />$x^2 + 4x + 4 = 8x + 4$<br />$x^2 - 4x = 0$<br />$x(x-4) = 0$<br /><br />So, the solutions are $x = 0$ and $x = 4$.<br /><br />Now, let's check these solutions by substituting them back into the original equation:<br /><br />For $x = 0$:<br />$\sqrt{3(0)+4} - \sqrt{2(0)+1} = 1$<br />$\sqrt{4} - \sqrt{1} = 1$<br />$2 - 1 = 1$<br />$1 = 1$ (True)<br /><br />For $x = 4$:<br />$\sqrt{3(4)+4} - \sqrt{2(4)+1} = 1$<br />$\sqrt{16} - \sqrt{9} = 1$<br />$4 - 3 = 1$<br />$1 = 1$ (True)<br /><br />Therefore, the solutions to the equation are $x = 0$ and $x = 4$. The correct answer is $\{0, 4\}$.