Problem 5 - A Graphical Approach to Precalculus with Limits

Question

These exercises review topics covered in earlier sections. The concepts are used in solving the applications that follow in this exercise set. Do not use a calculator. Solve \(-4 x^{2}+5 x=1\).

Step-by-Step Solution

Answer

The solutions are \(x = \frac{1}{4}\) and \(x = 1\).

Step 1: Set the equation to zero

Begin by rearranging the equation to set it equal to zero. Subtract 1 from both sides to give \[-4x^2 + 5x - 1 = 0.\]

Step 2: Apply the quadratic formula

Recall the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\] Identify the coefficients from the equation: \(a = -4\), \(b = 5\), \(c = -1\).

Step 3: Calculate the discriminant

The discriminant is \(b^2 - 4ac\). Substitute the values of \(a\), \(b\), and \(c\):\[b^2 - 4ac = 5^2 - 4(-4)(-1) = 25 - 16 = 9.\]

Step 4: Solve using the quadratic formula

Since the discriminant is a perfect square (9), continue with the quadratic formula:\[x = \frac{-5 \pm \sqrt{9}}{-8}.\]Simplify \(\sqrt{9}\) to obtain 3.

Step 5: Evaluate the expression

Substitute \(\pm 3\) into the expression and solve:\[x = \frac{-5 + 3}{-8} = \frac{-2}{-8} = \frac{1}{4},\]\[x = \frac{-5 - 3}{-8} = \frac{-8}{-8} = 1.\]

Step 6: Verify the solutions

Re-substitute the values \(x = \frac{1}{4}\) and \(x = 1\) into the original equation to ensure correctness. Both substitutions satisfy the equation.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool in algebra for solving quadratic equations. It is essential because it provides a general solution to any equation of the form \( ax^2 + bx + c = 0 \). Quadratic equations can describe various real-world phenomena, and understanding how to solve them is crucial.

The quadratic formula is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

To use this formula, follow the steps:

• Identify the coefficients \( a \), \( b \), and \( c \) from your quadratic equation.
• Substitute these values into the formula.
• Calculate the expression inside the square root, known as the discriminant.
• Solve for \( x \) using both the plus and minus variations of the \( \pm \) symbol.

This formula allows you to find the roots of the equation, which are the values for \( x \) that make the equation true.

Discriminant

The discriminant is a key part of the quadratic formula. It is represented by the expression \( b^2 - 4ac \) and can determine the nature and number of solutions for a quadratic equation.

Here's what you need to know about the discriminant:

• If the discriminant is positive, there are two distinct real roots.
• If the discriminant is zero, there is exactly one real root, also known as a repeated or double root.
• If the discriminant is negative, there are no real roots, but instead, two complex roots.

In our problem, the discriminant was calculated as 9, which is a perfect square and positive. This means we have two distinct real solutions for the equation. Understanding the discriminant helps you predict the nature of the roots even before explicitly solving the equation.

Solving Polynomial Equations

Solving polynomial equations involves finding the values of the variable that satisfy the equation. While linear equations are straightforward, polynomials like quadratic equations require additional methods.

For a quadratic equation:

• Start by setting the equation to zero. This standardizes the equation form to \( ax^2 + bx + c = 0 \).
• Use the quadratic formula or factorization, if applicable, to find solutions.
• Check the discriminant to determine the type and number of solutions.
• Remember to verify your solutions by substituting them back into the original equation.

Solving polynomial equations is a fundamental skill in algebra that lays the groundwork for more complex mathematical studies. With practice, the process becomes intuitive, allowing for more analytical and efficient problem-solving.