Solve the radical equation. Don t forget, you must check potential solutions. $5 \sqrt { x } = x + 4$
A) x = 16
B) x = 1
C) x = 1 or x = 16
D) x = $-$ 1 or x = 16

Question

Quizplus · Accepted Answer

First, we need to isolate the radical term by subtracting x from each side:

$$5\sqrt{x}-x=4$$

Next, we need to eliminate the radical by squaring both sides:

$$(5\sqrt{x}-x)^2=16$$

$$25x-10x\sqrt{x}+x^2=16$$

$$x^2-10x\sqrt{x}+25x-16=0$$

This is a quadratic equation in terms of $\sqrt{x}$, so we can let $y=\sqrt{x}$ and rewrite it as:

$$y^2-10y+25=0$$

$$(y-5)^2=0$$

$$y-5=0$$

$$y=5$$

So, $\sqrt{x}=5$ which means $x=25$. However, we must check this solution to make sure it doesn't result in any extraneous roots:

$$5\sqrt{25}=25+4$$

$$5\cdot 5=29$$

This is false, so $x=25$ is an extraneous root. Therefore, the only valid solution is $x=1$:

$$5\sqrt{1}=1+4$$

$$5=5$$

Which is true. Therefore, the answer is choice C: x = 1 or x = 16.

Solve the Radical Equation 5x=x+45 \sqrt { x } = x + 45x​=x+4

Solve the Radical Equation $5 \sqrt { x } = x + 4$