Problem 56 - Precalculus : Building Concepts and Connections

Question

Compute the zeros of the quadratic function. $$f(x)=2 x^{2}-x+8$$

Step-by-Step Solution

Answer

The roots of the quadratic function \(f(x)=2 x^{2}-x+8\) are \(0.25+1.98i\) and \(0.25-1.98i\).

Step 1: Identify the coefficients a, b, and c

From the quadratic function \(f(x) = 2x^{2} - x + 8\), we can identify the coefficients a, b, and c as follows: \(a=2\), \(b=-1\), and \(c=8\).

Step 2: Compute the discriminant

The discriminant is computed as \(b^{2} - 4ac = (-1)^{2} - 4*2*8 = 1 - 64 = -63\).

Step 3: Calculate the roots

Because the discriminant is negative, the quadratic function has no real roots. The roots are complex and can be found using the quadratic formula \(\frac{-b±\sqrt{b^{2} - 4ac}}{2a}\). Let's consolidate this as -b is 1, 2a is 4 and \(\sqrt{b^{2} - 4ac}\) is \(\sqrt{63}i\) which is approximated to 7.94i. Hence the roots are \(x=\frac{1±7.94i}{4}\), or \(x=0.25±1.98i\).

Key Concepts

Discriminant in Quadratic Equations

The discriminant is a critical component when working with quadratic equations, such as \( ax^2 + bx + c = 0 \). The formula for the discriminant \( D \) is given by \( b^2 - 4ac \). It tells you the nature of the roots of the quadratic equation.

When you compute the discriminant:

• If \( D > 0 \), there are two distinct real roots.
• If \( D = 0 \), there is exactly one real root (a repeated root).
• If \( D < 0 \), there are no real roots; instead, the roots are complex.

In our original exercise example, we found \( D = -63 \). Since \( -63 < 0 \), it indicates that the quadratic equation does not have real roots but rather complex roots.

Complex Roots

When the discriminant is negative, quadratic equations have complex roots. These roots are not on the real plane but involve imaginary numbers. Imaginary numbers are essential in mathematics, and their core component is \( i \), where \( i^2 = -1 \).

In the context of our quadratic equation \( 2x^2 - x + 8 \), the negative discriminant \( -63 \) results in complex roots. The quadratic formula helps compute these roots in the form \( \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

The square root of a negative discriminant, like \( \sqrt{-63} \), involves \( i \). This becomes \( \sqrt{63}i \), where \( \sqrt{63} \) is approximately \( 7.94 \). Hence, the roots can be represented as complex numbers, specifically:

• \( x = 0.25 + 1.98i \)
• \( x = 0.25 - 1.98i \)

These physical numbers have both a real part (\( 0.25 \)) and an imaginary part (\( 1.98i \)).

Quadratic Formula

The quadratic formula is a universal method for solving quadratic equations. Whether the roots are real or complex, it can always be applied to determine them. The formula is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

To use the quadratic formula:

• Identify the coefficients \( a \), \( b \), and \( c \) from the quadratic equation.
• Calculate the discriminant \( b^2 - 4ac \).
• Plug these values into the quadratic formula to get the roots.

In the exercise example, after computing the discriminant as \(-63\), which is negative, using the formula helped identify complex roots. Remember that the \( \pm \sqrt{b^2 - 4ac} \) part gives you two solutions: one with addition and the other with subtraction. These results confirmed the roots as \( x = 0.25 \pm 1.98i \). The quadratic formula thereby demonstrates its utility in managing both real and imaginary numbers seamlessly.