Problem 46 - Precalculus : Building Concepts and Connections

Question

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

Step-by-Step Solution

Answer

The real solutions to the equation are \(t = -2\) and \(t = 1/2\).

Step 1: Identify the coefficients

In the given quadratic equation, \(-2 + t^{2} + t = 0\), we can rewrite it in the standard form as \(t^{2}+t-2 = 0\). So we can identify the coefficients as \(a = 1\), \(b = 1\) and \(c = -2\).

Step 2: Determine the determinant

Calculate the discriminant, which is \(b^{2}-4ac\). Thus, discriminant equals to \(1^{2}-4(1)(-2)= 1+8=9\).

Step 3: Apply the Quadratic Formula

Plugging the coefficients \(a\), \(b\) and \(c\) into the quadratic formula, solving it would give the solutions to the equation. This gives us \(t = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} = \frac{-1 \pm \sqrt{9}}{2} = -1 \pm \frac{3}{2}\). This simplifies to \(t = -2\) or \(t = 1/2\).

Key Concepts

Solving Quadratic Equations

Quadratic equations are often presented in the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \), represent known values and \( a \) is not equal to zero. The solutions to these equations, also known as 'roots', are the values of \( x \) that make the equation true. The widely recognized method for solving these equations is the quadratic formula: \[ x = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{{2a}} \]

To solve these equations, three main steps are commonly followed: identifying coefficients, calculating the discriminant, and then applying the Quadratic Formula. By following these steps, we can ensure that we find the most accurate solutions possible, whether they are real numbers or complex numbers. It's essential to perform each step carefully to avoid errors that might lead to incorrect solutions.

Discriminant in Quadratics

The discriminant of a quadratic equation, symbolized by \( D \), is a part of the quadratic formula and is given by the expression \( b^2 - 4ac \). It holds a crucial role in determining the nature and number of the solutions of the equation.

• If \( D > 0 \), the quadratic equation has two distinct real solutions.
• If \( D = 0 \), the equation has exactly one real solution.
• If \( D < 0 \), the equation has two complex solutions and no real solutions.

In our example, the discriminant is \( 9 \), which is positive, hence the equation \( t^2 + t - 2 = 0 \) has two distinct real solutions. Understanding the discriminant is key to predicting the number and type of solutions without actually solving the equation.

Real Solutions of Quadratics

Real solutions of a quadratic equation are the \( x \) values that satisfy the equation when plotted on a real number line. It's possible when the discriminant \( D \), \( b^2 - 4ac \), is greater than or equal to zero, as nonnegative discriminants indicate real number solutions.

Following the quadratic formula and the discriminant, we determine the real solutions by performing operations included in the formula. In our exercise, we found two real solutions \( t = -2 \) and \( t = 1/2 \). These solutions are the points where the graph of the quadratic equation \( t^2 + t - 2 = 0 \) intersects with the \( t \) axis. This concept is essential as understanding real solutions helps us analyze and interpret the behavior of quadratic functions in practical scenarios.