Question

Find the real solutions, if any, of each equation. Use the quadratic formula. $$ 2 x^{2}=1-2 x $$

Short answer

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

Step-by-step solution

Rewrite the equation in standard form

The given equation is: \[ 2x^2 = 1 - 2x \]First, move all terms to one side of the equation to get it into the standard quadratic form \(ax^2 + bx + c = 0\).Subtract 1 and add 2x to both sides:\[ 2x^2 + 2x - 1 = 0 \]

Identify coefficients

Identify the coefficients \(a\), \(b\), and \(c\) from the standard form equation \(2x^2 + 2x - 1 = 0\).Here, \(a = 2\), \(b = 2\), and \(c = -1\).

Write down the quadratic formula

The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Substitute the coefficients into the formula

Substitute \(a = 2\), \(b = 2\), and \(c = -1\) into the quadratic formula:\[ x = \frac{-(2) \pm \sqrt{(2)^2 - 4(2)(-1)}}{2(2)} \]

Simplify the expression inside the square root

Simplify the expression under the square root first:\[ \sqrt{(2)^2 - 4(2)(-1)} = \sqrt{4 + 8} = \sqrt{12} = 2\sqrt{3} \]

Complete the quadratic formula calculation

Substitute back and simplify the expression:\[ x = \frac{-2 \pm 2\sqrt{3}}{4} \]Factor out a 2 in the numerator:\[ x = \frac{2(-1 \pm \sqrt{3})}{4} = \frac{-1 \pm \sqrt{3}}{2} \]

State the solutions

The solutions to the quadratic equation are:\[ x = \frac{-1 + \sqrt{3}}{2} \text{ and } x = \frac{-1 - \sqrt{3}}{2} \]

Key concepts

Solving Quadratic Equations

Quadratic equations are a fundamental part of algebra and appear in various applications ranging from physics to finance. To solve a quadratic equation means to find the values of the variable that make the equation true.
If you have an equation in the form of \(ax^2 + bx + c = 0\), you can use different methods to find the solutions, also called roots. These methods include:

• Factoring
• Completing the Square
• Using the Quadratic Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Each method has its own advantages, but the Quadratic Formula is a universally applicable method. It works for any quadratic equation as long as you correctly identify the coefficients \(a\), \(b\), and \(c\). Let's look deeper into the specific parts of a quadratic equation to understand how to solve them effectively.

Standard Form of Quadratic Equation

Before solving a quadratic equation, it's essential to express it in its standard form: \(ax^2 + bx + c = 0\). This form allows you to easily identify the coefficients \(a\), \(b\), and \(c\).

Take the equation given in the exercise: \(2x^2 = 1 - 2x\). To get it into standard form, follow these steps:

• Move all terms to one side of the equation: Subtract 1 and add 2x to both sides to get: \(2x^2 + 2x - 1 = 0\).
• Identify the coefficients:
• \(a = 2\)
• \(b = 2\)
• \(c = -1\)

Once in this form, you can proceed to solve the equation using any suitable method, such as the Quadratic Formula.

Roots of Quadratic Equation

The roots of a quadratic equation are the values of \(x\) that satisfy the equation \(ax^2 + bx + c = 0\). These roots can be real or complex numbers. In this exercise, we focus on finding the real roots using the Quadratic Formula.

From our example, we substitute the identified coefficients into the quadratic formula:

\[x = \frac{-(2) \pm \sqrt{(2)^2 - 4(2)(-1)}}{2(2)}\]

Simplify under the square root:
\[\sqrt{4 + 8} = \sqrt{12} = 2\sqrt{3}\]

Finally, simplify the entire expression:
\[x = \frac{-2 \pm 2\sqrt{3}}{4} = \frac{-1 \pm \sqrt{3}}{2}\]

This gives us two real roots:
\[x = \frac{-1 + \sqrt{3}}{2} \text{ and } x = \frac{-1 - \sqrt{3}}{2}\]

These roots are the solutions to the quadratic equation \(2x^2 + 2x - 1 = 0\).