Problem 15 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation. For equations with real solutions, support your answers graphically. $$2 x^{2}=90$$

Step-by-Step Solution

Answer

The solutions are \( x = 3\sqrt{5} \) and \( x = -3\sqrt{5} \).

Step 1: Write the Equation

Start by writing down the equation that we need to solve: \[ 2x^2 = 90 \]

Step 2: Isolate the Squared Term

To simplify the equation, divide both sides by 2 to isolate the squared term on one side: \[ x^2 = \frac{90}{2} = 45 \]

Step 3: Solve for x

Take the square root of both sides to solve for \( x \): \[ x = \pm \sqrt{45} \] Simplify \( \sqrt{45} \) by breaking it into prime factors: \[ \sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5} \] Thus, the solutions are: \[ x = 3\sqrt{5} \quad \text{or} \quad x = -3\sqrt{5} \]

Step 4: Graphical Representation

To graphically represent the solution, sketch a graph of \( y = 2x^2 \), which is a parabola opening upwards. The line \( y = 90 \) is a horizontal line. Since the solutions to the equation are where \( 2x^2 = 90 \), these are the points where the parabola intersects the line \( y = 90 \). These intersection points correspond to \( x = 3\sqrt{5} \) and \( x = -3\sqrt{5} \).

Key Concepts

Real Solutions

Quadratic equations often have what's called "real solutions." These are the x-values where the quadratic equation equals zero or intersects a horizontal line when visualized on a graph.
In the equation \( 2x^2 = 90 \), to solve for real solutions, we isolate the x-term. This includes isolating the square term \( x^2 \) and then taking the square root to determine the potential values of \( x \).

Let's break down the important steps to find real solutions:

• Isolating the Squared Term: We need to manipulate the equation until the squared term, \( x^2 \), stands alone. Hence, divide both sides by 2 to achieve \( x^2 = 45 \).
• Taking the Square Root: By taking the square root of 45, we determine \( x = \pm \sqrt{45} \). This gives us two solutions because a square root can be both positive and negative.

Thus, the real solutions for \( x \) are \( 3\sqrt{5} \) and \( -3\sqrt{5} \). Real solutions are the actual numeric values that satisfy the initial equation.

Graphical Representation

Graphing is an excellent tool to understand where solutions of equations lie. For the quadratic function \( y = 2x^2 \), we can visualize its shape and how it relates to the solutions.
When graphed, a quadratic equation like \( y = 2x^2 \) appears as a curve known as a parabola.

Here’s how this works in practice:

• Understanding the Shape: A parabola opens upwards or downwards depending on the coefficient of \( x^2 \). Here, \( 2 \) is positive, so the parabola opens up.
• Finding Intersections: The real solutions occur where this parabola intersects the line \( y = 90 \). By plotting both \( y = 2x^2 \) and \( y = 90 \), you find intersection points at \( x = 3\sqrt{5} \) and \( x = -3\sqrt{5} \).

Graphical representation defines not just solutions but also illustrates their position and behavior visually on a coordinate plane.

Squaring and Square Roots

For any quadratic equation, understanding the technique of squaring and finding roots is critical. This process helps solve quadratic equations by determining the potential values of \( x \) that meet the equation's requirements.
Let's delve into the related mathematical steps:

• Squaring a Term: Initially, we see terms like \( 2x^2 \), which is a multiplication of 2 by the square of \( x \), showcasing the typical form of a quadratic equation.
• Taking Square Roots: After isolating \( x^2 = 45 \), we derive \( x \) by taking the square root. This step is key to transitioning from squared terms back to basic x-values: \( x = \pm \sqrt{45} = \pm 3\sqrt{5} \).
• Simplifying Radicals: Expressing square roots in their simplest form aids in clarity. For \( \sqrt{45} \), we look for factors like \( \sqrt{9 \times 5} \), leading to \( \sqrt{9} \times \sqrt{5} = 3\sqrt{5} \).

This technique reveals critical solutions and reinforces how square roots directly undo squaring, allowing us to work backward from equations to find x-values.