Chapter 3: Problem 5
Solve each problem. Do not use a calculator. $$\text { Solve }-4 x^{2}+5 x=1$$
Short Answer
Expert verified
The solutions are \(x = \frac{1}{4}\) and \(x = 1\).
Step by step solution
01
Set the Equation to Zero
Start by rearranging the equation \,\(-4x^2 + 5x = 1\)\, into a standard quadratic equation form, which is \,\(ax^2 + bx + c = 0\). To do this, subtract \(1\) from both sides to obtain: \,\(-4x^2 + 5x - 1 = 0\).
02
Identify Coefficients
Identify the coefficients from the quadratic equation \(-4x^2 + 5x - 1 = 0\). Here, \(a = -4\), \(b = 5\), and \(c = -1\).
03
Use the Quadratic Formula
The quadratic formula is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute \(a = -4\), \(b = 5\), and \(c = -1\) into the formula to find the solutions for \(x\).
04
Calculate the Discriminant
First, calculate the discriminant \(b^2 - 4ac\). Substitute the values: \(\text{Discriminant} = 5^2 - 4(-4)(-1) = 25 - 16 = 9\). Since the discriminant is positive, there are two real solutions.
05
Solve using the Quadratic Formula
Substitute the discriminant back into the quadratic formula: \[x = \frac{-5 \pm \sqrt{9}}{2(-4)}\] Now solve for each scenario of \(\pm\).
06
Solve for \(x_1\)
Calculate \(x_1\) using the plus sign: \[x_1 = \frac{-5 + \sqrt{9}}{-8} = \frac{-5 + 3}{-8} = \frac{-2}{-8} = \frac{1}{4}\] Thus, \(x_1 = \frac{1}{4}\).
07
Solve for \(x_2\)
Calculate \(x_2\) using the minus sign: \[x_2 = \frac{-5 - \sqrt{9}}{-8} = \frac{-5 - 3}{-8} = \frac{-8}{-8} = 1\] Thus, \(x_2 = 1\).
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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 reliable method for solving any quadratic equation, which is in the form of \(ax^2 + bx + c = 0\). This formula can be very helpful, especially when other methods like factoring are not possible. The formula is as follows: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] It allows you to find the values of \(x\) that satisfy the equation. You simply plug in the coefficients \(a\), \(b\), and \(c\) from the quadratic equation into the formula.
- \(b\): the coefficient of \(x\)
- \(a\): the coefficient of \(x^2\)
- \(c\): the constant term
Discriminant
The discriminant plays a key role in determining the nature of the solutions of a quadratic equation. It is the part of the quadratic formula under the square root symbol: \(b^2 - 4ac\). Calculating the discriminant is a valuable step because it tells us about the number and type of solutions without solving the equation fully. Here’s how it works:
- If the discriminant is positive (\(b^2 - 4ac > 0\)): you will have two distinct real solutions. The parabola crosses the x-axis at two points.
- If the discriminant is zero (\(b^2 - 4ac = 0\)): there is exactly one real solution. The parabola touches the x-axis at one point.
- If the discriminant is negative (\(b^2 - 4ac < 0\)): there are no real solutions. The solutions are complex or imaginary, meaning the parabola does not touch the x-axis.
Real Solutions
Real solutions are solutions to a quadratic equation that can be plotted on a graph as points where the parabola touches or crosses the x-axis. Recognizing if the solutions of an equation are real or not can greatly impact how you interpret the findings of any problem involving quadratic equations.
- Two Real Solutions: Occurs when the discriminant is positive. Each solution corresponds to a point where the parabola intersects the x-axis.
- One Real Solution: Happens when the discriminant is zero, indicating the parabola just touches the x-axis.
- No Real Solutions: A negative discriminant means no intersection with the x-axis; solutions are complex numbers.