Question Solve the radical equation $\sqrt{5 x+50}=x$ , and check all proposed solutions.

Studdy Solution

STEP 1

Assumptions
1. We are given the radical equation $\sqrt{5x + 50} = x$ .
2. We need to solve for $x$ .
3. We must check all proposed solutions to ensure they are valid within the original equation.

STEP 2

To solve the radical equation, we will first isolate the radical on one side and then square both sides of the equation to eliminate the square root.
$\left(\sqrt{5x + 50}\right)^2 = x^2$

STEP 3

Squaring both sides of the equation gives us:
$(5x + 50) = x^2$

STEP 4

Now, we need to set the equation to zero by moving all terms to one side to form a quadratic equation.
$x^2 - 5x - 50 = 0$

STEP 5

Next, we factor the quadratic equation if possible. We look for two numbers that multiply to $-50$ and add to $-5$ .

STEP 6

The numbers that satisfy these conditions are $-10$ and $5$ . We can now factor the quadratic equation as follows:
$(x - 10)(x + 5) = 0$

STEP 7

Using the zero-product property, we set each factor equal to zero and solve for $x$ .
$x - 10 = 0 \quad \text{or} \quad x + 5 = 0$

STEP 8

Solving the first equation for $x$ gives us:
$x = 10$

STEP 9

Solving the second equation for $x$ gives us:
$x = -5$

STEP 10

We have found two potential solutions, $x = 10$ and $x = -5$ . However, we must check both solutions in the original radical equation to ensure they are valid.

STEP 11

First, we check $x = 10$ :
$\sqrt{5(10) + 50} = 10$

STEP 12

Simplify the expression inside the square root:
$\sqrt{50 + 50} = 10$

STEP 13

Further simplify the expression:
$\sqrt{100} = 10$

STEP 14

Since the square root of $100$ is indeed $10$ , the equation holds true:
$10 = 10$

STEP 15

Thus, $x = 10$ is a valid solution.

STEP 16

Now, we check $x = -5$ :
$\sqrt{5(-5) + 50} = -5$

STEP 17

Simplify the expression inside the square root:
$\sqrt{-25 + 50} = -5$

STEP 18

Further simplify the expression:
$\sqrt{25} = -5$

STEP 19

Since the square root of $25$ is $5$ and not $-5$ , the equation does not hold true:
$5 \neq -5$

STEP 20

Thus, $x = -5$ is not a valid solution because the square root of a number is always non-negative.

STEP 21

The only valid solution to the original radical equation is $x = 10$ .
The correct answer is A. $\{10\}$ .

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