Question

cubic and cube root functions and equations
what is the extraneous solution to the equation $sqrt{2p + 1} + 2sqrt{p} = 1$
$p = 0$
$p = -2$
$p = 4$
$p = -4$

Explanation:

Step1: Isolate the radical term

$$\sqrt{2p+1}=1-2\sqrt{p}$$

Step2: Square both sides

$$(\sqrt{2p+1})^2=(1-2\sqrt{p})^2$$
$$2p+1=1-4\sqrt{p}+4p$$

Step3: Simplify the equation

$$2p+1-1-4p=-4\sqrt{p}$$
$$-2p=-4\sqrt{p}$$
$$p=2\sqrt{p}$$

Step4: Square both sides again

$$p^2=(2\sqrt{p})^2$$
$$p^2=4p$$
$$p^2-4p=0$$
$$p(p-4)=0$$
$$p=0 \text{ or } p=4$$

Step5: Check solutions in original equation

For $p=0$:
$$\sqrt{2(0)+1}+2\sqrt{0}=1+0=1 \quad \text{(valid)}$$
For $p=4$:

$$\sqrt{2(4)+1}+2\sqrt{4}=\sqrt{9}+2(2)=3+4=7 eq1 \quad \text{(extraneous)}$$

For $p=-2$ and $p=-4$: $\sqrt{p}$ is not real, so they are not valid candidates.

Answer:

$p=4$