Question

solving equations containing two radicals
solving equations containing two radicals
which statement about $sqrt4{2x} + sqrt4{x + 3} = 0$ is true?
$x = 3$ is a true solution.
$x = 3$ is an extraneous solution.
$x = 1$ is a true solution.
$x = 1$ is an extraneous solution.

Explanation:

Step1: Isolate one radical

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

Step2: Raise both sides to 4th power

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

Step3: Solve for x

$2x - x = 3$
$x = 3$

Step4: Verify the solution

Substitute $x=3$ into original equation:
$\sqrt[4]{2(3)} + \sqrt[4]{3+3} = \sqrt[4]{6} + \sqrt[4]{6} = 2\sqrt[4]{6}
eq 0$
Since the left side does not equal 0, $x=3$ is extraneous.

Step5: Check x=1 (for completeness)

Substitute $x=1$ into original equation:
$\sqrt[4]{2(1)} + \sqrt[4]{1+3} = \sqrt[4]{2} + \sqrt[4]{4} = \sqrt[4]{2} + \sqrt{2}
eq 0$
So $x=1$ is not a solution.

Answer:

$x = 3$ is an extraneous solution.