Problem 46 - Calculus with Concepts in Calculus

Question

Approximate all zeros of the function to the nearest hundredth. $$ f(x)=\sqrt{2} x^{2}+\pi x+1 $$

Step-by-Step Solution

Answer

The zeros are approximately \(-0.39\) and \(-1.84\).

Step 1: Identify the function type

The function given is a quadratic function of the form \( ax^2 + bx + c \). Here, \( a = \sqrt{2} \), \( b = \pi \), and \( c = 1 \). Quadratic functions can have up to two real zeros.

Step 2: State the quadratic formula

The zeros of a quadratic function \( ax^2 + bx + c = 0 \) can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

Step 3: Compute the discriminant

Calculate the discriminant using the formula \( b^2 - 4ac \). Substitute the values \( b = \pi \), \( a = \sqrt{2} \), and \( c = 1 \): \[ b^2 - 4ac = \pi^2 - 4(\sqrt{2})(1) \].

Step 4: Simplify the discriminant

Calculate \( \pi^2 \approx 3.14159^2 \approx 9.8696 \) and \( 4\sqrt{2} = 4 \times 1.4142 \approx 5.6568 \). Thus, the discriminant is \( 9.8696 - 5.6568 = 4.2128 \).

Step 5: Calculate the zeros

Since the discriminant is positive, there are two real zeros. Substitute into the quadratic formula: \[ x = \frac{-\pi \pm \sqrt{4.2128}}{2\sqrt{2}} \]. Next, calculate \( \sqrt{4.2128} \approx 2.0521 \).

Step 6: Calculate each zero

Calculate the zeros individually. First, \[ x_1 = \frac{-\pi + 2.0521}{2\sqrt{2}} = \frac{-3.1416 + 2.0521}{2\times 1.4142} \approx \frac{-1.0895}{2.8284} \approx -0.385 \]. Then for the second zero, \[ x_2 = \frac{-\pi - 2.0521}{2\sqrt{2}} = \frac{-3.1416 - 2.0521}{2.8284} \approx \frac{-5.1937}{2.8284} \approx -1.837 \].

Step 7: Round to the nearest hundredth

Round the calculated zeros to the nearest hundredth: \( x_1 \approx -0.39 \) and \( x_2 \approx -1.84 \).

Key Concepts

Quadratic Formula

The quadratic formula is a magic tool for solving any quadratic equation. These equations take the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. This formula gives you the values of \( x \) that make the equation true, called the roots or zeros of the equation.
To find these roots, you use the formula:

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

Here, the "\( \pm \)" symbol means there will typically be two solutions: one for the plus and one for the minus.
In our problem, the quadratic formula helps us discover where the function \( f(x) = \sqrt{2} x^2 + \pi x + 1 \) touches or crosses the x-axis.
Using our formula, everything relies on finding the right numbers for \( a \), \( b \), and \( c \), then clever arithmetic to reach our solution!

Discriminant

The discriminant is an integral part of the quadratic formula. It helps to determine the nature of the roots without solving the full equation. The expression for the discriminant is:

• \[ b^2 - 4ac \]

What does this value tell us?
- If the discriminant is positive, like it is in our problem (computed to be approximately 4.2128), the quadratic equation has two distinct real zeros.
- If it's zero, there is exactly one real root, meaning the parabola just kisses the x-axis.
- If the discriminant is negative, there are no real roots, meaning the solutions are complex numbers.
In the given exercise, calculating \( b^2 \) and \(- 4ac \) carefully shows our equation's two real crossings on the x-axis. This helps confirm the calculations and provides assurance that there are realistic solutions to find using the quadratic formula.

Real Zeros

Real zeros are the x-values where the y-value of a function becomes zero. In easier terms, these are points where the function crosses or touches the x-axis.
To find the real zeros, we rely on the well-discussed quadratic formula and the discriminant. Once we apply these tools, we solve for \( x \) using:

• \[ x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \]
• \[ x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} \]

From the example's careful computations, we found two zeros: \( x_1 \approx -0.39 \) and \( x_2 \approx -1.84 \). These zeros are rounded to the nearest hundredth for precision.Breaking it down, these numbers mean that at \( x = -0.39 \) and \( x = -1.84 \), the function's value becomes zero! This conclusion tells us exactly where our parabola intersects the x-axis, giving us useful insights into the graph's behavior.