Problem 15 - Beginning and Intermediate Algebra

Question

Use the quadratic formula to solve each equation. (All solutions for these equations are real numbers.) $$ 2 x^{2}-2 x=1 $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{1 + \sqrt{3}}{2}\) and \(x = \frac{1 - \sqrt{3}}{2}\).

Step 1 - Write the equation in standard form

First, rewrite the equation in the standard form of a quadratic equation, which is \(ax^2 + bx + c = 0\). Starting with the given equation: \[2x^2 - 2x = 1\]

Step 2 - Move all terms to one side

Subtract 1 from both sides of the equation to set it to zero: \[2x^2 - 2x - 1 = 0\]

Step 3 - Identify coefficients

Identify the coefficients \(a\), \(b\), and \(c\) in the standard form equation: - \(a = 2\)- \(b = -2\)- \(c = -1\)

Step 4 - Use the quadratic formula

The quadratic formula is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Substituting the coefficients into the quadratic formula: \[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(-1)}}{2(2)}\]

Step 5 - Simplify the expression

Simplify under the square root and the fraction: First calculate the discriminant: \[(-2)^2 - 4(2)(-1) = 4 + 8 = 12\] Then plug back in: \[x = \frac{2 \pm \sqrt{12}}{4}\] Simplify further by breaking down \(\sqrt{12}\): \[x = \frac{2 \pm 2\sqrt{3}}{4} = \frac{2(1 \pm \sqrt{3})}{4} = \frac{1 \pm \sqrt{3}}{2}\]

Key Concepts

quadratic formula

The quadratic formula is an essential tool for solving quadratic equations. A quadratic equation is any equation that can be written in the form:
\[ ax^2 + bx + c = 0 \]
To find the solutions (or roots) of such an equation, we use the quadratic formula:
\[ x = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \]
This formula helps you find the values of \( x \) that make the equation true by plugging in the coefficients \( a \), \( b \), and \( c \).
Here are the key steps to using the quadratic formula:

• Identify the coefficients \( a \), \( b \), and \( c \) from your quadratic equation.
• Plug these coefficients into the quadratic formula.
• Calculate the discriminant \( b^2 - 4ac \) within the formula.
• Simplify the results to find the values of \( x \).

Let's see it with an example from the exercise: Given:
\[ 2x^2 - 2x - 1 = 0 \]
We find \( a = 2 \), \( b = -2 \), and \( c = -1 \). Then, plug them into the formula:
\[ x = \frac{-(-2) \, \pm \, \sqrt{(-2)^2 - 4(2)(-1)}}{2(2)} \]
This gives us two potential values for \( x \).

standard form

Before using the quadratic formula, you must ensure your quadratic equation is in standard form:
\[ ax^2 + bx + c = 0 \]
In this form, the equation is set equal to zero with all terms on one side.
For example, if we start with \[2x^2 - 2x = 1 \], we need to move the 1 to the left side to get:
\[2x^2 - 2x - 1 = 0 \]
Now, the equation is in standard form, making it ready for using the quadratic formula.
If the given quadratic equation is not in standard form, here’s what to do:

• Move all terms to one side of the equation so that it equals zero.
• Combine like terms and simplify where necessary.
• Ensure the equation follows the structure of \(ax^2 + bx + c = 0\).

This form is crucial for easily identifying the coefficients \(a\), \(b\), and \(c\).

discriminant

The discriminant is a part of the quadratic formula located under the square root symbol:
\[ b^2 - 4ac \]
The value of the discriminant determines the nature of the roots:

• 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, only complex roots.

In the exercise, we calculate the discriminant:
Starting with \( 2x^2 - 2x - 1 = 0 \), we identify \( a = 2 \), \( b = -2 \), and \( c = -1 \).
Plugging these into the discriminant formula:
\[ (-2)^2 - 4(2)(-1) = 4 + 8 = 12 \]
Since the discriminant (12) is greater than zero, we have two distinct real roots.
This tells us we will find two different values for \( x \), confirming they are real and distinct.