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Magazine Chapter 6: Problem 68
Solve each of the following equations for \(x\). $$x^{2}+8 a x+15 a^{2}=0$$
Short Answer
Expert verified
The solutions are \(x = -3a\) and \(x = -5a\).
Step by step solution
01
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 \(c = 15a^2\).
02
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 \(c = 15a^2\).
03
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: \[\Delta = (8a)^2 - 4 \times 1 \times 15a^2 = 64a^2 - 60a^2 = 4a^2\].
04
Compute the Square Root of the Discriminant
The square root of the discriminant \(\sqrt{\Delta}\) is \(\sqrt{4a^2}\). Calculate it as \(2a\).
05
Solve for x Using the Formula
Substitute the discriminant and other values into the quadratic formula: \[x = \frac{-(8a) \pm 2a}{2 \times 1} = \frac{-8a \pm 2a}{2}\].
06
Simplify the Solutions
Calculate each solution separately: \(x_1 = \frac{-8a + 2a}{2} = \frac{-6a}{2} = -3a\) and \(x_2 = \frac{-8a - 2a}{2} = \frac{-10a}{2} = -5a\).
07
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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).
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鈥檚 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.
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).
