Solve.
-$\sqrt { 2 x + 3 } + \sqrt { 4 - x } = 4$
A) $- 3$
B) $3 , \frac { 11 } { 9 }$
C) 3
D) No solution

Question

Quizplus · Accepted Answer

To solve the equation $\sqrt { 2 x + 3 } + \sqrt { 4 - x } = 4$, we first isolate one of the square roots and then square both sides to eliminate it. Let's isolate $\sqrt { 2 x + 3 }$:$\sqrt { 2 x + 3 } = 4 - \sqrt { 4 - x }$Squaring both sides gives:$2x + 3 = 16 - 8\sqrt { 4 - x } + 4 - x$Simplifying:$3x - 10 = -8\sqrt { 4 - x }$Dividing by -8:$\frac{3x - 10}{-8} = \sqrt { 4 - x }$Squaring both sides again:$\left(\frac{3x - 10}{-8}\right)^2 = 4 - x$Solving this equation will give us the potential solutions. However, because we squared the equation twice, we must check each solution in the original equation to ensure it doesn't introduce extraneous solutions. Solving the quadratic equation and checking the solutions, we find that $x = 3$ and $x = \frac{11}{9}$ satisfy the original equation. Therefore, the correct answer is B) $3 , \frac { 11 } { 9 }$.

Solve - 2x+3+4−x=4\sqrt { 2 x + 3 } + \sqrt { 4 - x } = 42x+3​+4−x​=4

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