solve the radical equation, and check all proposed solutions. 36) $sqrt{2x + 3}-sqrt{x + 1}=1$

Explanation:

Step1: Isolate one radical

Isolate $\sqrt{2x + 3}$: $\sqrt{2x+3}=1 + \sqrt{x + 1}$

Step2: Square both sides

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

Step3: Simplify the equation

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

Step4: Square both sides again

$(x + 1)^2=(2\sqrt{x + 1})^2$ $x^{2}+2x + 1 = 4(x + 1)$ $x^{2}+2x+1=4x + 4$ $x^{2}+2x-4x+1 - 4=0$ $x^{2}-2x - 3=0$

Step5: Factor the quadratic equation

$(x - 3)(x+ 1)=0$ So $x=3$ or $x=-1$

Step6: Check the solutions

For $x = 3$: Left - hand side: $\sqrt{2\times3+3}-\sqrt{3 + 1}=\sqrt{9}-\sqrt{4}=3 - 2=1$, which is equal to the right - hand side. For $x=-1$: Left - hand side: $\sqrt{2\times(-1)+3}-\sqrt{-1 + 1}=\sqrt{1}-0 = 1$, which is equal to the right - hand side.

Answer:

$x = 3$ or $x=-1$