Problem 26 - Algebra for College Students

Question

Solve each radical equation. Don't forget, you must check potential solutions. $$\sqrt{5 x}+10=x$$

Step-by-Step Solution

Answer

The solution is \( x = 20 \).

Step 1: Isolate the Radical Expression

The equation is given as \( \sqrt{5x} + 10 = x \). First, isolate the radical expression by subtracting 10 from both sides: \[ \sqrt{5x} = x - 10 \]

Step 2: Square Both Sides

To eliminate the square root, square both sides of the equation: \[ (\sqrt{5x})^2 = (x - 10)^2 \]Simplifying this, we get: \[ 5x = x^2 - 20x + 100 \]

Step 3: Reorganize the Equation

Rearrange the terms to form a quadratic equation: \[ 0 = x^2 - 25x + 100 \]This can also be written as: \[ x^2 - 25x + 100 = 0 \]

Step 4: Solve the Quadratic Equation

The quadratic equation \( x^2 - 25x + 100 = 0 \) can be factored. To factor, find two numbers that multiply to 100 and add to -25. These numbers are -20 and -5, so we factor the equation as:\[ (x-20)(x-5) = 0 \]Setting each factor equal to zero gives potential solutions:\[ x - 20 = 0 \] or \[ x - 5 = 0 \]Thus, \( x = 20 \) or \( x = 5 \).

Step 5: Check Potential Solutions

Substitute each potential solution back into the original equation to see if it holds true.- Check for \( x = 20 \): \( \sqrt{5 \times 20} + 10 = 20 \) \( \sqrt{100} + 10 = 20 \) \( 10 + 10 = 20 \) which is true.- Check for \( x = 5 \): \( \sqrt{5 \times 5} + 10 = 5 \) \( \sqrt{25} + 10 = 5 \) \( 5 + 10 = 5 \) which is false.

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two. They typically take the form: \[x^2 + bx + c = 0\]Here, \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). When solving quadratic equations derived from radical equations, as in our exercise, the transition involves eliminating the radical by squaring both sides, which can introduce these second-degree equations.

In our example, by rearranging and simplifying the equation after isolating the radical term, we arrived at the quadratic equation:\[x^2 - 25x + 100 = 0\]Recognizing that an equation is quadratic is essential because it informs us on how to approach solving it, which we can often do through factoring.

Factoring

Factoring is a method used to solve quadratic equations by expressing the quadratic expression as a product of its linear factors. This process simplifies solving the equation by identifying solutions that make each factor equal to zero.

For instance, let's factor the quadratic equation from our example:\[x^2 - 25x + 100 = 0\]We look for two numbers that multiply to 100 and sum up to -25. These numbers are -20 and -5.

• Multiply: \(-20 \times -5 = 100\)
• Add: \(-20 + -5 = -25\)

Clever factoring rewrites the equation as:\[(x-20)(x-5) = 0\] Factoring is powerful because once completed, it allows us to set each factor equal to zero to find the possible values of \(x\).

Solving Equations

Solving the factored equation involves setting each factor to zero and solving for \(x\).

In our case, the solutions are found like this:

• \(x - 20 = 0\): Solving gives \(x = 20\)
• \(x - 5 = 0\): Solving gives \(x = 5\)

These solutions represent the values that satisfy the factored quadratic equation individually. This step is crucial as it provides potential answers that we have set to check against the original equation.

Checking Solutions

Once we have potential solutions from solving the quadratic equation, the final step is verification. Substituting back into the original equation helps ensure these solutions are valid, not artifacts introduced by the operations (like squaring).

Let's check them:

• For \(x = 20\):
  In the original equation \(\sqrt{5x} + 10 = x\), substitute \(x = 20\): \[\sqrt{5 \times 20} + 10 = 20\] Simplifies to: \[10 + 10 = 20\] which is correct.
• For \(x = 5\):
  Substitute \(x = 5\): \[\sqrt{5 \times 5} + 10 = 5\] Simplifies to: \[5 + 10 = 5\] which is incorrect.

Verification is key as it confirms that \(x = 20\) is the only true solution, ensuring alignment with our original radical equation.