Problem 40 - College Algebra with Modeling and Visualization

Question

Solve the equation. Check your answers. $$ \sqrt{4-3 x}=x+8 $$

Step-by-Step Solution

Answer

The solution is \( x = -4 \).

Step 1: Isolate the Square Root

The equation is \( \sqrt{4-3x} = x + 8 \). Since we have a square root on one side of the equation, an effective method is to isolate this square root on one side, which has already been done for us: \( \sqrt{4-3x} \).

Step 2: Square Both Sides

To eliminate the square root, square both sides of the equation. This gives us:\[ (\sqrt{4-3x})^2 = (x + 8)^2 \]Simplifying this, we get:\[ 4 - 3x = x^2 + 16x + 64 \]

Step 3: Rearrange the Equation

Bring all terms to one side to set the equation to zero:\[ x^2 + 16x + 64 - 4 + 3x = 0 \]Simplifying, this equation becomes:\[ x^2 + 19x + 60 = 0 \]

Step 4: Solve the Quadratic Equation

To solve the quadratic equation \( x^2 + 19x + 60 = 0 \), use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1 \), \( b = 19 \), and \( c = 60 \). Substitute these values into the formula:\[ x = \frac{-19 \pm \sqrt{19^2 - 4 \times 1 \times 60}}{2 \times 1} \]\[ x = \frac{-19 \pm \sqrt{361 - 240}}{2} \]\[ x = \frac{-19 \pm \sqrt{121}}{2} \]\[ x = \frac{-19 \pm 11}{2} \]This results in two solutions:\[ x = \frac{-19 + 11}{2} = -4 \]and\[ x = \frac{-19 - 11}{2} = -15 \]

Step 5: Check the Solutions

Substitute \( x = -4 \) and \( x = -15 \) back into the original equation to check:1. For \( x = -4 \):\[ \sqrt{4 - 3(-4)} = -4 + 8 \]\[ \sqrt{16} = 4 \] which holds true as \( 4 = 4 \).2. For \( x = -15 \):\[ \sqrt{4 - 3(-15)} = -15 + 8 \]\[ \sqrt{49} = -7 \] which is false as \( 7 eq -7 \).Thus, only \( x = -4 \) is a valid solution.

Key Concepts

Square Roots

Square roots are a mathematical concept where you determine a number that, when multiplied by itself, gives the original number. In our exercise, we have a square root on one side of the equation, \( \sqrt{4-3x} \). To solve equations involving square roots, it's crucial to isolate the square root before proceeding with further steps. This clarifies which expression is the root, making it easier to eliminate it by squaring. Remember:

• Isolate the square root if it's not already done.
• Ensure the expression under the square root is non-negative, as square root is not defined for negative numbers in real numbers.
• Squaring both sides of an equation is a common method to eliminate the square root.

Practicing these steps helps build confident problem-solving skills related to roots and radicals.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). In our scenario, after rearranging and simplifying, we end up with the quadratic equation \( x^2 + 19x + 60 = 0 \). The formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]To effectively use this formula:

• Identify coefficients: \( a \), \( b \), and \( c \) from the equation.
• Substitute these values into the formula.
• Calculate the discriminant, \( b^2 - 4ac \), which determines the number and type of solutions.
• Find the two possible values for \( x \) by using the plus and minus signs in the formula.

The quadratic formula always provides solutions when they exist, even if they are complex numbers.

Equation Solving

Solving equations means finding the value of the variable that satisfies the equation. In our exercise, we began by isolating the square root and then eliminating it by squaring both sides. This led us to a more common type of equation: the quadratic equation. The process involves:

• Isolating terms where necessary.
• Transforming the equation by performing operations like squaring.
• Rearranging terms to simplify and set the equation to zero.
• Solving the quadratic equation using methods like factoring or quadratic formula.

Each step is crucial for reaching the correct solution, and understanding these techniques helps in managing different types of equations.

Mathematical Solutions

Mathematical solutions require verifying that the solutions satisfy the original conditions of the equation. After finding potential solutions, substituting these back into the initial equation confirms their validity. In our exercise:

• We found two potential solutions: \( x = -4 \) and \( x = -15 \).
• Checking \( x = -4 \) shows it does satisfy the original equation.
• Checking \( x = -15 \) reveals it does not satisfy the equation, hence it's not a valid solution.

This step is important to eliminate extraneous solutions, especially when equations involve operations like squaring, which can introduce false solutions. Always complete this step to ensure accuracy.