Problem 68 - Algebra for College Students

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: Identify the Equation Type

The given equation \(x^2 + 8ax + 15a^2 = 0\) is a quadratic equation. A quadratic equation is generally of the form \(ax^2 + bx + c = 0\). In this case: \(a = 1\), \(b = 8a\), and \(c = 15a^2\).

Step 2: Apply the Quadratic Formula

For a quadratic equation \(ax^2 + bx + c = 0\), the solutions for \(x\) are given by the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. In this equation, substitute \(a = 1\), \(b = 8a\), and \(c = 15a^2\).

Step 3: Calculate the Discriminant

The discriminant \(\Delta\) is given by \(b^2 - 4ac\). Substitute the values: \(b = 8a\), \(a = 1\), and \(c = 15a^2\) to find: \[\Delta = (8a)^2 - 4 \times 1 \times 15a^2 = 64a^2 - 60a^2 = 4a^2\].

Step 4: Compute the Square Root of the Discriminant

The square root of the discriminant \(\sqrt{\Delta}\) is \(\sqrt{4a^2}\). Calculate it as \(2a\).

Step 5: Solve for x Using the Formula

Substitute the discriminant and other values into the quadratic formula: \[x = \frac{-(8a) \pm 2a}{2 \times 1} = \frac{-8a \pm 2a}{2}\].

Step 6: Simplify the Solutions

Calculate each solution separately: \(x_1 = \frac{-8a + 2a}{2} = \frac{-6a}{2} = -3a\) and \(x_2 = \frac{-8a - 2a}{2} = \frac{-10a}{2} = -5a\).

Step 7: Verify the Solutions

To ensure correctness, substitute each solution back into the original equation. For \(x = -3a\): \((-3a)^2 + 8a(-3a) + 15a^2 = 9a^2 - 24a^2 + 15a^2 = 0\). For \(x = -5a\): \((-5a)^2 + 8a(-5a) + 15a^2 = 25a^2 - 40a^2 + 15a^2 = 0\). Both solutions satisfy the initial equation.

Key Concepts

Quadratic Formula

The quadratic formula is a crucial tool for solving quadratic equations. A quadratic equation is any equation expressed in the form: \[ ax^2 + bx + c = 0 \]. It may look complex, but with the quadratic formula, solving it becomes manageable. The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. This formula helps find the values of \(x\) that make the equation true. Here's a quick breakdown of the components:

• \(a\), \(b\), and \(c\) are coefficients in the quadratic equation.
• \(b^2 - 4ac\) is called the discriminant, determining the nature of the roots.
• \(\pm\) indicates that there are typically two possible solutions (roots).

Apply the formula by substituting the values of \(a\), \(b\), and \(c\) from your specific equation. This will yield the possible solutions for \(x\). The quadratic formula is versatile; it works for all quadratic equations, whether they factor neatly or not.

Discriminant

The discriminant in a quadratic equation offers key insights into the nature of its solutions. It is expressed as part of the quadratic formula: \[ b^2 - 4ac \], and it tells us how many and what kind of roots the equation has.Consider these scenarios:

• If the discriminant is positive, \(b^2 - 4ac > 0\), the equation has two distinct real roots.
• If it is zero, \(b^2 - 4ac = 0\), there’s exactly one real solution, also called a repeated or double root.
• If it is negative, \(b^2 - 4ac < 0\), no real number solutions exist; the roots are complex numbers.

Calculating the discriminant is a great way to save time, as it tells you upfront what to expect in terms of solutions. In our specific case, substituting the real-world values into the formula helps determine the types of numbers you’re dealing with (real or complex), thereby guiding the solution strategy.

Solving Equations

Solving quadratic equations might seem daunting at first, but breaking it down into logical steps makes it simpler. The method to solve involves:

• Identifying the equation as a quadratic, which generally looks like \(ax^2 + bx + c = 0\).
• Using the quadratic formula: plugging in the coefficients to find the roots (solutions) of the equation.
• Calculating the discriminant, \(b^2 - 4ac\), to understand the nature of the roots before proceeding to find their exact values.
• Applying the formula thoroughly, substituting your found discriminant and coefficients, and simplifying the results to derive two potential \(x\) values (or note if one/two real solutions exist).

After solving, it's wise to validate your solutions by plugging them back into the original equation. This ensures that no errors occurred throughout calculations.