Problem 82 - College Algebra with Modeling and Visualization

Question

Complete the following. (a) Write the equation as \(a x^{2}+b x+c=0\) with \(a>0\) (b) Calculate the discriminant \(b^{2}-4 a c\) and determine the number of real solutions. (c) Solve the equation. $$ 8 x^{2}-2=14 $$

Step-by-Step Solution

Answer

The solutions are \(x = \pm \sqrt{2}\).

Step 1: Rewrite the Equation in Standard Form

First, we need to rewrite the given equation \(8x^2 - 2 = 14\) in the form \(ax^2 + bx + c = 0\). To do this, subtract 14 from both sides to get:\[ 8x^2 - 2 - 14 = 0 \]This simplifies to:\[ 8x^2 - 16 = 0 \]Now, the equation is in standard form with \(a=8\), \(b=0\), and \(c=-16\).

Step 2: Calculate the Discriminant

The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by the formula:\[ b^2 - 4ac \]Substitute \(a=8\), \(b=0\), and \(c=-16\) into the formula:\[ \Delta = 0^2 - 4 \times 8 \times (-16) \]\[ \Delta = 0 + 512 = 512 \]

Step 3: Determine the Number of Real Solutions

The discriminant helps determine the number of real solutions:- If \(\Delta > 0\), there are two distinct real solutions.- If \(\Delta = 0\), there is one real solution.- If \(\Delta < 0\), there are no real solutions.Since \(\Delta = 512 > 0\), there are two distinct real solutions.

Step 4: Solve the Quadratic Equation

We will use the quadratic formula to solve:\[ x = \frac{{-b \pm \sqrt{\Delta}}}{{2a}} \]We know \(b=0\), so:\[ x = \frac{{0 \pm \sqrt{512}}}{{2 \times 8}} \]Simplify:\[ x = \frac{{\pm \sqrt{512}}}{16} \]We find \(\sqrt{512}\):\(\sqrt{512} = \sqrt{256 \times 2} = 16\sqrt{2}\)So:\[ x = \frac{{\pm 16\sqrt{2}}}{16} \]Finally, simplify to find:\[ x = \pm \sqrt{2} \]

Key Concepts

Discriminant

The discriminant is a crucial element when it comes to understanding quadratic equations. It helps us predict the nature and number of solutions without solving the entire equation. The discriminant is denoted by the Greek letter \(\Delta\) and is calculated using the formula:

• \[ \Delta = b^2 - 4ac \]

For our quadratic equation \(8x^2 - 16 = 0\), we have \(a = 8\), \(b = 0\), and \(c = -16\). By plugging these values into the formula, we compute the discriminant:

• \[ \Delta = 0^2 - 4 \times 8 \times (-16) = 512 \]

A positive discriminant, as in this case where \(\Delta = 512\), indicates that there are two distinct real solutions. If the discriminant were zero, it would mean one real solution, and if it were negative, it would mean no real solutions, only complex solutions.

Real Solutions

Real solutions are values of \(x\) that satisfy the equation and result in a zero when plugged back into the equation. The discriminant is key in determining the number of these solutions:

• If \(\Delta > 0\), there are two distinct real solutions.
• If \(\Delta = 0\), there is exactly one real solution, often called a repeated or double root.
• If \(\Delta < 0\), there are no real solutions; instead, the solutions will be complex numbers.

For the equation \(8x^2 - 16 = 0\), the discriminant was found to be 512, which is greater than zero, confirming that there are two different real solutions. These real solutions can be calculated using further methods, predominantly the quadratic formula.

Quadratic Formula

The quadratic formula is a robust tool for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). It enables us to find the solutions, or roots, of the quadratic equation. The formula is:

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

This formula incorporates all necessary components: \(a\), \(b\), \(c\), and the discriminant \(b^2 - 4ac\). Let's apply it to our specific equation where \(a = 8\), \(b = 0\), and \(c = -16\):

• \[ x = \frac{0 \pm \sqrt{512}}{16} \]

To simplify \(\sqrt{512}\), we find that \(\sqrt{512} = 16\sqrt{2}\). Plugging this back, we have:

• \[ x = \frac{\pm 16\sqrt{2}}{16} = \pm \sqrt{2} \]

Thus, the solutions are \(x = \sqrt{2}\) and \(x = -\sqrt{2}\). These are the two real solutions we confirmed using the discriminant in the previous sections.