Problem 41 - Precalculus : Building Concepts and Connections

Question

Solve the quadratic equation by using the quadratic formula. Find only real solutions. $$x^{2}+2 x-1=0$$

Step-by-Step Solution

Answer

The real solutions to the equation \(x^{2} + 2x - 1 = 0\) are \(x=-1+\sqrt{2}\) and \(x=-1-\sqrt{2}\).

Step 1 - Identify the coefficients

The equation is given by \(x^{2}+2x-1=0\). Here, the coefficients are \(a=1\), \(b=2\), \(c=-1\)

Step 2 - Substitute the coefficients into the quadratic formula

Now that we have the coefficients, let's substitute them into the quadratic formula, which is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). We get: \(x = \frac{-2 \pm \sqrt{(2^2) - 4*(1)*(-1)}}{2*1}\)

Step 3 - Simplify within the square root

Simplify within the square root to obtain the discriminant. It turns out that \(b^{2}-4ac=(2^2)-4*(1)*(-1)=4+4=8\).

Step 4 - Simplify the equation

With the discriminant calculated, it is now possible to simplify the expression to \(x = \frac{-2 \pm \sqrt{8}}{2}\). The square root of 8 can be simplified to \(2\sqrt{2}\). Thus, the simplification gives \(x = \frac{-2 \pm 2\sqrt{2}}{2}\). When dividing everything by 2, the final equation turns into \(x = -1 \pm \sqrt{2}\)

Step 5 - Find the solutions

This equation presents two solutions, namely \(x=-1+\sqrt{2}\) and \(x=-1-\sqrt{2}\)

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are in the form \(ax^2 + bx + c = 0\). To find the solutions, you can use the formula:

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

Here,

• \(a\), \(b\), and \(c\) are coefficients from the equation.
• \(b^2 - 4ac\) is known as the discriminant, which we'll discuss in detail shortly.

To use the quadratic formula, simply plug in the values of \(a\), \(b\), and \(c\) that come from the quadratic equation. Place them into the formula, and work through simplifying the calculations.
The formula gives us two potential solutions because of the \(\pm\). It implies you'll need to perform two calculations: once with a plus sign and once with a minus sign.
This formula allows you to find exact calculations easily, even for complex equations.

Real Solutions

When solving a quadratic equation, such as \(x^2 + 2x - 1 = 0\), finding real solutions means identifying values of \(x\) that are real numbers. Some quadratic solutions can be imaginary, depending on the discriminant's value.
In our context of finding real solutions, it is crucial to:

• Evaluate the discriminant \(b^2 - 4ac\). If the value is positive or zero, real solutions exist.
• Note that a positive discriminant results in two distinct real solutions, while a zero discriminant yields one repeated real solution.
• If the discriminant is negative, the equation has no real solutions, only complex ones.

For the exercise \(x^2 + 2x - 1 = 0\), by calculating, we found that the discriminant is 8. Since 8 is positive, it confirms the existence of two distinct real solutions, \(x = -1 + \sqrt{2}\) and \(x = -1 - \sqrt{2}\).

Discriminant

The discriminant, key to understanding the nature of solutions in a quadratic equation, is represented by the expression \(b^2 - 4ac\). It informs us about the nature and the number of solutions a quadratic equation has.

• If \(b^2 - 4ac > 0\), the quadratic equation has two distinct real solutions.
• If \(b^2 - 4ac = 0\), the equation has exactly one real solution (also known as a repeated root).
• If \(b^2 - 4ac < 0\), there are no real solutions, but rather two complex solutions.

In the example \(x^2 + 2x - 1 = 0\), we calculated the discriminant as \(b^2 - 4ac = 8\). This value tells us that two real solutions exist.
Understanding the discriminant helps you quickly categorize the solutions. This makes the process smooth and efficient, whether dealing with real or complex numbers.