Problem 83 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ x=\frac{\sqrt{12 x-5}}{2} $$

Step-by-Step Solution

Answer

The solutions are \(x = 2.5\) and \(x = 0.5\). Both are valid.

Step 1: Eliminate the Fraction

Multiply both sides of the equation by 2 to eliminate the fraction:\[ 2x = \sqrt{12x - 5} \]

Step 2: Square Both Sides

To eliminate the square root, square both sides of the equation:\[ (2x)^2 = (\sqrt{12x - 5})^2 \]This simplifies to:\[ 4x^2 = 12x - 5 \]

Step 3: Rearrange the Equation

Rearrange the equation to set it to zero:\[ 4x^2 - 12x + 5 = 0 \]

Step 4: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -12\), and \(c = 5\), to solve for \(x\):\[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4\cdot4\cdot5}}{2\cdot4} \]Calculate the discriminant:\[ (-12)^2 - 4 \cdot 4 \cdot 5 = 144 - 80 = 64 \]Now solve:\[ x = \frac{12 \pm \sqrt{64}}{8} \]\[ x = \frac{12 \pm 8}{8} \]This results in two solutions:\[ x = \frac{12 + 8}{8} = 2.5 \]\[ x = \frac{12 - 8}{8} = 0.5 \]

Step 5: Check for Extraneous Solutions

Substitute each solution back into the original equation to check for validity.For \(x = 2.5\):\[ 2.5 = \frac{\sqrt{12\cdot2.5 - 5}}{2} \]Calculate inside the square root:\[ 12\cdot2.5 - 5 = 30 - 5 = 25 \]\[ \frac{\sqrt{25}}{2} = \frac{5}{2} = 2.5 \] (valid)For \(x = 0.5\):\[ 0.5 = \frac{\sqrt{12\cdot0.5 - 5}}{2} \]Calculate inside the square root:\[ 12\cdot0.5 - 5 = 6 - 5 = 1 \]\[ \frac{\sqrt{1}}{2} = \frac{1}{2} = 0.5 \] (valid)

Step 6: Conclusion

Both solutions \(x = 2.5\) and \(x = 0.5\) hold true when substituted back into the original equation, thus neither solution is extraneous.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). This formula provides a straightforward way to find the roots (solutions) of a quadratic equation. The formula is expressed as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• \( a \) is the coefficient of \( x^2 \).
• \( b \) is the coefficient of \( x \).
• \( c \) is the constant term.

The term under the square root, \( b^2 - 4ac \), is called the discriminant. It tells us about the nature of the roots:

• If it's positive, the equation has two distinct real roots.
• If it's zero, there is exactly one real root.
• If it's negative, the roots are complex numbers.

In our exercise, after simplifying, the equation is \( 4x^2 - 12x + 5 = 0 \). Plugging \( a = 4 \), \( b = -12 \), and \( c = 5 \) into the formula gives us the solutions \( x = 2.5 \) and \( x = 0.5 \). Here, the discriminant was positive (64), indicating two real solutions.

Extraneous Solutions

When solving radical equations or equations involving fractions, it's crucial to be aware of extraneous solutions. An extraneous solution is a solution that emerges from the algebraic process but doesn't satisfy the original equation.This often occurs when both sides of an equation are squared to eliminate a square root. Squaring can introduce new solutions that never existed in the original equation.
To identify extraneous solutions, each proposed solution must be substituted back into the original equation to verify its validity.
In the presented problem, both potential solutions \( x = 2.5 \) and \( x = 0.5 \) were verified. Substituting these values back into the original equation \( x = \frac{\sqrt{12x - 5}}{2} \) confirmed that both were true solutions, meaning there were no extraneous solutions in this instance.

Checking Solutions in Algebra

Checking solutions is a critical step in solving equations to ensure that the solutions found are correct, especially when solving equations through methods that can introduce extraneous solutions, like squaring.Here's how to check solutions in algebra, using our previous example:

• Take each solution and substitute it back into the original equation.
• Solve the equation to see if both sides are equal.

For instance, substituting \( x = 2.5 \) into the original equation \( x = \frac{\sqrt{12x - 5}}{2} \):

• Plug \( 2.5 \) in and simplify: \( 2.5 = \frac{\sqrt{12 \cdot 2.5 - 5}}{2} \).
• Calculate: \( 12 \cdot 2.5 - 5 = 25 \).
• Find the square root: \( \sqrt{25} = 5 \).
• Then divide: \( \frac{5}{2} = 2.5 \). Confirming it balances the equation.

Repeat for \( x = 0.5 \):

• Substitute and simplify: \( 0.5 = \frac{\sqrt{12 \cdot 0.5 - 5}}{2} \).
• Calculate: \( 12 \cdot 0.5 - 5 = 1 \).
• Find the square root: \( \sqrt{1} = 1 \).
• Then divide: \( \frac{1}{2} = 0.5 \). It checks out as valid.

This process confirms the accuracy of both solutions, proving they satisfy the original equation.