Problem 44 - Elementary and Intermediate Algebra

Question

Solve each equation. Let \(f(x)=\sqrt{2 x^{2}-7 x}\). For what value(s) of \(x\) is \(f(x)=2 ?\)

Step-by-Step Solution

Answer

The value of \(x\) for which \(f(x)=2\) is \(-\frac{1}{2}\).

Step 1: Set Up the Equation

Given that \(f(x) = \sqrt{2x^2 - 7x}\), we need to find when \(f(x) = 2\). Set the equation \(\sqrt{2x^2 - 7x} = 2\).

Step 2: Eliminate the Square Root

Square both sides of the equation to remove the square root: \((\sqrt{2x^2 - 7x})^2 = 2^2\). This simplifies to \(2x^2 - 7x = 4\).

Step 3: Rearrange into Standard Form

Move all terms to one side of the equation to arrange it as a standard quadratic equation: \(2x^2 - 7x - 4 = 0\).

Step 4: Solve the Quadratic Equation

Use the quadratic formula to solve for \(x\). The formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -7\), and \(c = -4\).

Step 5: Calculate the Discriminant

First, calculate the discriminant: \(b^2 - 4ac = (-7)^2 - 4 \cdot 2 \cdot (-4) = 49 + 32 = 81\).

Step 6: Apply the Quadratic Formula

Since the discriminant is 81, which is a perfect square, apply the quadratic formula: \(x = \frac{-(-7) \pm \sqrt{81}}{2 \cdot 2} = \frac{7 \pm 9}{4}\).

Step 7: Find the Possible Values of x

Calculate the roots: \(x = \frac{7 + 9}{4} = 4\) and \(x = \frac{7 - 9}{4} = -\frac{1}{2}\).

Step 8: Verify the Solutions

Check both values to ensure they solve the original equation: For \(x = 4\), \(f(4) = \sqrt{2(4)^2 - 7(4)} = \sqrt{16} = 4 eq 2\). For \(x = -\frac{1}{2}\), \(f(-\frac{1}{2}) = \sqrt{2(-\frac{1}{2})^2 - 7(-\frac{1}{2})} = \sqrt{\frac{1}{2} + \frac{7}{2}} = \sqrt{4} = 2\). Only \(x = -\frac{1}{2}\) is valid.

Key Concepts

Quadratic Formula

When solving quadratic equations, the quadratic formula is your best friend. It gives you a clear path to finding the roots of any quadratic equation. The formula itself is:

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

Here, \(a\), \(b\), and \(c\) are coefficients from your standard quadratic equation \(ax^2 + bx + c = 0\). This formula works by providing solutions (or "roots") where the equation equals zero. These are the points where the parabola, a graphical representation of the quadratic equation, intersects the x-axis.
Using the quadratic formula becomes much easier with practice. Just remember to carefully identify the coefficients \(a\), \(b\), and \(c\), and insert them carefully into the formula. This method is foolproof and widely applicable, no matter how complex the quadratic equation might seem.

Discriminant

The discriminant is a special part of the quadratic formula: \(b^2 - 4ac\). It's a crucial piece because it tells us about the nature of the roots of the quadratic equation.
Here's what the discriminant reveals:

• If \(b^2 - 4ac > 0\), there are two distinct real roots.
• If \(b^2 - 4ac = 0\), there is exactly one real root (a repeated root).
• If \(b^2 - 4ac < 0\), there are no real roots; instead, there are two complex roots.

This tiny piece of the formula can save you time by letting you know what type of solutions to expect before you delve into calculations. In the example we examined, the discriminant was 81, a positive perfect square, indicating the presence of two distinct real roots. This helps you plan your approach towards solving the equation with confidence.

Functions

Functions are a foundational concept in mathematics. Simply put, a function is a relationship between a set of inputs and a set of possible outputs, typically represented as \(f(x)\).
A function assigns exactly one output to each valid input. In the problem we're looking at, \(f(x) = \sqrt{2x^2 - 7x}\) was defined, and we were tasked with finding for what value(s) of \(x\) this function equals 2.

• Functions can represent many types of relationships, not just those described by quadratic expressions.
• Understanding how to manipulate and solve functions is crucial because they model real-world scenarios throughout science, engineering, and everyday life.

Grasping the underlying relationship between input values (\(x\)) and output values (\(f(x)\)) will make working with functions intuitive and straightforward.

Algebraic Expressions

Algebraic expressions form the backbone of many mathematical problems. They are combinations of numbers, variables, and arithmetic operations.
Working with these expressions involves:

• Identifying terms: These could be constants, such as \(-7\), or terms like \(2x^2\) which feature variables.
• Simplifying expressions: Combine like terms to make the expression more manageable.
• Transforming equations: Use algebraic techniques to move terms to either side of the equation, which is crucial for solving equations like the quadratic ones.

In the example given, the expression \(2x^2 - 7x - 4 = 0\) needed careful handling. Understanding each component of the expression makes it possible to use powerful tools like the quadratic formula effectively.
Mastering algebraic expressions unlocks pathways to solving more complex mathematical challenges.