Problem 54 - Algebra and Trigonometry

Question

Use the quadratic formula and a calculator to find all real solutions, rounded to three decimals. $$ x^{2}-2.450 x+1.500=0 $$

Step-by-Step Solution

Answer

The real solutions are \(x = 1.250\) and \(x = 1.200\).

Step 1: Understanding the Problem

We need to solve the quadratic equation \(x^2 - 2.450x + 1.500 = 0\) using the quadratic formula to find all the real solutions. The quadratic formula is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation \(ax^2 + bx + c = 0\).

Step 2: Identify Coefficients

From the equation \(x^2 - 2.450x + 1.500 = 0\), identify the coefficients as follows: \(a = 1\), \(b = -2.450\), and \(c = 1.500\).

Step 3: Calculate the Discriminant

Calculate the discriminant using the formula \(b^2 - 4ac\). Substitute the values to get \((-2.450)^2 - 4 \times 1 \times 1.500 = 6.0025 - 6 = 0.0025\).

Step 4: Apply the Quadratic Formula

Now apply the quadratic formula: \[ x = \frac{-(-2.450) \pm \sqrt{0.0025}}{2 \times 1} \]Simplify the expression: \( x = \frac{2.450 \pm 0.05}{2} \).

Step 5: Calculate the Solutions

Calculate the two possible values of \(x\):1. \(x = \frac{2.450 + 0.05}{2} = 1.250\)2. \(x = \frac{2.450 - 0.05}{2} = 1.200\)

Step 6: Round the Solutions

Both solutions \(x = 1.250\) and \(x = 1.200\) are already rounded to three decimal places.

Key Concepts

Real Solutions

Real solutions in a quadratic equation are values of \(x\) that make the equation true and belong to the set of real numbers. These solutions are determined by the points where a parabola, represented by the equation, intersects the x-axis.

• If the parabola touches the x-axis at one point, the equation has one real solution, also known as a repeated root.
• If it intersects at two points, there are two distinct real solutions.
• When the parabola does not touch the x-axis at all, there are no real solutions.

In our specific exercise, the equation \(x^2 - 2.450x + 1.500 = 0\) has two real solutions, \(x = 1.250\) and \(x = 1.200\), indicating two distinct points of intersection on the x-axis.

Quadratic Equation

A quadratic equation represents a polynomial equation of degree two, and it can typically be expressed in the standard form as \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable that we're solving for.

• If \(a = 0\), the equation is not quadratic because without \(x^2\), it reduces to a linear equation.
• The graph of a quadratic equation is a parabola, which opens upwards if \(a > 0\) and downwards if \(a < 0\).

Recognizing the coefficients is crucial when using the quadratic formula. In our equation, \(a = 1\), \(b = -2.450\), and \(c = 1.500\). The careful identification of these values allows us to further explore roots using the quadratic formula.

Discriminant

The discriminant is a component of the quadratic formula, denoted as \(b^2 - 4ac\). It helps determine the nature and quantity of solutions of the quadratic equation.

• If the discriminant is positive, the equation has two distinct real solutions.
• If it is zero, there is exactly one real solution, which means the parabola touches the x-axis at one point.
• If it is negative, the equation has no real solutions as the parabola does not intersect the x-axis.

In this exercise, the discriminant \(0.0025\) was calculated, which is a small positive number. Although very close to zero, it implies that there are indeed two real solutions, slightly differing from each other. This demonstrates the utility of the discriminant in predicting the number of real solutions for a quadratic equation.