Problem 68 - Intermediate Algebra

Question

Solve each of the following equations for \(x\). $$ x^{2}+8 a x+15 a^{2}=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = -3a \) and \( x = -5a \).

Step 1: Write Down the Quadratic Equation

First, identify and write down the given quadratic equation, which is \( x^2 + 8ax + 15a^2 = 0 \). This equation is in the standard form \( ax^2 + bx + c = 0 \).

Step 2: Identify Coefficients

Identify the coefficients in the quadratic equation. Here, the coefficient of \( x^2 \) is 1, the coefficient of \( x \) is \( 8a \), and the constant term is \( 15a^2 \).

Step 3: Apply the Quadratic Formula

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( x \). In this context, \( a = 1 \), \( b = 8a \), and \( c = 15a^2 \).

Step 4: Calculate the Discriminant

Find the discriminant \( \Delta = b^2 - 4ac \). Substitute the values to get \( \Delta = (8a)^2 - 4(1)(15a^2) = 64a^2 - 60a^2 = 4a^2 \).

Step 5: Solve for x Using the Quadratic Formula

Substitute the values into the quadratic formula: \[ x = \frac{-(8a) \pm \sqrt{4a^2}}{2(1)} = \frac{-8a \pm 2a}{2} \].

Step 6: Simplify the Expression

Simplify the expression for \( x \) to find the two possible values: \[ x = \frac{-8a + 2a}{2} = \frac{-6a}{2} = -3a \] and \[ x = \frac{-8a - 2a}{2} = \frac{-10a}{2} = -5a \].

Step 7: Write the Solutions

The solutions to the quadratic equation are \( x = -3a \) and \( x = -5a \).

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations, which are in the form \( ax^2 + bx + c = 0 \). The solution for \( x \) can be found using the formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula might seem complicated at first, but it’s just a matter of plugging in the coefficients from your equation. For example, in the equation \( x^2 + 8ax + 15a^2 = 0 \):

• \( a \) is 1
• \( b \) is \( 8a \)
• \( c \) is \( 15a^2 \)

By substituting these coefficients into the formula, you can solve for \( x \). This process will give you two solutions due to the \( \pm \) in the equation, which reflects the parabola's symmetrical nature in its graph representation.

Discriminant

The discriminant is an important part of the quadratic formula. It's the expression under the square root in the formula: \( b^2 - 4ac \). The discriminant helps determine the number and type of solutions of a quadratic equation.

• If the discriminant is positive, there are two distinct real solutions.
• If it's zero, there is one repeated real solution.
• If it's negative, the solutions are complex or imaginary.

For the exercise in question, the discriminant is calculated as: \[\Delta = (8a)^2 - 4 \times 1 \times 15a^2 = 64a^2 - 60a^2 = 4a^2 \] Since \( 4a^2 \) is positive, the quadratic equation \( x^2 + 8ax + 15a^2 = 0 \) has two real and distinct solutions. Understanding the discriminant helps predict the outcome even before solving the entire equation.

Factoring Quadratic Equations

Factoring is another method to solve quadratic equations. Sometimes, a quadratic equation can be factored into the product of two binomials, making it easier to find solutions. Consider the process and try to express a quadratic equation like \( x^2 + 8ax + 15a^2 \) in a factored form like \((x + m)(x + n) = 0\). If such a factorization is possible, the solutions are simply \( x = -m \) and \( x = -n \). For this exercise, notice that:

• The sum of \( m \) and \( n \) would be \(-8a \)
• The product of \( m \) and \( n \) would be \( 15a^2 \)

Using these clues, \(-3a\) and \(-5a\) are the correct factors because:

• Their sum is indeed \(-8a\).
• Their product is \(15a^2\).

So, the original equation can be rewritten as \((x + 3a)(x + 5a) = 0\), leading to solutions \( x = -3a \) and \( x = -5a \). Factoring is particularly useful for neat and simple equations.