Chapter 6: Problem 59
Solve each equation. See Examples I through 9. (A few exercises are linear equations.) $$ 5 x^{2}-6 x-8=0 $$
Short Answer
Expert verified
The solutions are \( x = 2 \) and \( x = -\frac{4}{5} \).
Step by step solution
01
Identify the equation type
First, we identify the given equation as a quadratic equation. A quadratic equation can be recognized by having the form \( ax^2 + bx + c = 0 \). Here, the given equation is \( 5x^2 - 6x - 8 = 0 \), which fits the quadratic form with \( a = 5 \), \( b = -6 \), and \( c = -8 \).
02
Apply the quadratic formula
To solve the quadratic equation \( 5x^2 - 6x - 8 = 0 \), we'll use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Substituting \( a = 5 \), \( b = -6 \), and \( c = -8 \) into the formula, we get: \[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 5 \cdot (-8)}}{2 \cdot 5} \].
03
Calculate the discriminant and simplify
The discriminant \( b^2 - 4ac \) is a part of the quadratic formula that determines the nature of the roots. Let's calculate it: \((-6)^2 - 4 \cdot 5 \cdot (-8) = 36 + 160 = 196\). Now substitute this back into the quadratic formula: \[ x = \frac{6 \pm \sqrt{196}}{10} \]. Since \( \sqrt{196} = 14 \), the expression becomes: \[ x = \frac{6 \pm 14}{10} \].
04
Solve for the roots
We now solve for both possible values of \( x \) using the results from the previous step. First calculate \( x = \frac{6 + 14}{10} = \frac{20}{10} = 2 \). Next, calculate the second option: \( x = \frac{6 - 14}{10} = \frac{-8}{10} = -\frac{4}{5} \).
05
State the solutions
The solutions to the quadratic equation \( 5x^2 - 6x - 8 = 0 \) are \( x = 2 \) and \( x = -\frac{4}{5} \). These represent the values of \( x \) for which the equation is satisfied.
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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 powerful tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). This formula provides a straightforward method to find the solutions, or "roots," of a quadratic equation without needing to factor it. Here is the quadratic formula in its standard form: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]When using the quadratic formula:
- Identify the coefficients \(a\), \(b\), and \(c\) in the quadratic equation.
- Substitute these values into the formula.
- Remember that \( \pm \) means you will likely find two solutions: one with a plus and one with a minus.
Discriminant
The discriminant is a component of the quadratic formula within the square root: \( b^2 - 4ac \). It plays a critical role in determining the number and type of roots a quadratic equation has. Calculating the discriminant can tell us whether the roots are real or imaginary, and if they are distinct or repeated.Here's what the discriminant indicates:
- If \( b^2 - 4ac > 0 \), the quadratic equation has two distinct real roots.
- If \( b^2 - 4ac = 0 \), there is exactly one real root, also known as a repeated or double root.
- If \( b^2 - 4ac < 0 \), the equation has no real roots; instead, it has two complex roots.
Roots of Equation
The roots of a quadratic equation are the values of \( x \) that make the equation true (i.e., when the equation equals zero). Finding the roots is the ultimate goal of solving a quadratic equation since they indicate where the function graph intersects the x-axis.Based on the quadratic formula, the roots obtained for the equation \( 5x^2 - 6x - 8 = 0 \) are:
- The first root: \( x = 2 \)
- The second root: \( x = -\frac{4}{5} \)