Problem 26 - Algebra for College Students

Question

Use the quadratic formula to solve each of the quadratic equations. Check your solutions by using the sum and product relationships. $$3 x^{2}-2 x+5=0$$

Step-by-Step Solution

Answer

The quadratic equation has complex solutions \( x = \frac{1 + i\sqrt{14}}{3} \) and \( x = \frac{1 - i\sqrt{14}}{3} \).

Step 1: Identify coefficients

Identify the coefficients \( a \), \( b \), and \( c \) in the quadratic equation. The given equation is \( 3x^2 - 2x + 5 = 0 \). Here, \( a = 3 \), \( b = -2 \), and \( c = 5 \).

Step 2: Write the quadratic formula

Recall the quadratic formula for finding the roots of the equation \( ax^2 + bx + c = 0 \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Step 3: Calculate the discriminant

Calculate the discriminant \( \Delta = b^2 - 4ac \). Substitute our values into the formula: \[ (-2)^2 - 4(3)(5) = 4 - 60 = -56 \] The discriminant is \(-56\), which is less than zero, indicating that there are no real roots, only complex ones.

Step 4: Express the complex roots

Since the discriminant is negative, the roots will be complex numbers. Substitute into the quadratic formula: \[ x = \frac{2 \pm \sqrt{-56}}{6} \] Simplify the expression using \( \sqrt{-56} = i\sqrt{56} \), where \( i = \sqrt{-1} \).

Step 5: Simplify the complex roots

Further simplify the square root part: \[ \sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14} \] Thus, the roots are \[ x = \frac{2 \pm 2i\sqrt{14}}{6} = \frac{1 \pm i\sqrt{14}}{3} \] These are the solutions: \( x = \frac{1 + i\sqrt{14}}{3} \) and \( x = \frac{1 - i\sqrt{14}}{3} \).

Step 6: Check the solutions using sum and product relationships

For a quadratic equation \( ax^2 + bx + c = 0 \), the sum of roots \( \alpha + \beta = -\frac{b}{a} \) and the product \( \alpha \beta = \frac{c}{a} \). Thus: Sum: \( \frac{1 + i\sqrt{14}}{3} + \frac{1 - i\sqrt{14}}{3} = \frac{2}{3} \) Product: \( \left( \frac{1 + i\sqrt{14}}{3} \right) \left( \frac{1 - i\sqrt{14}}{3} \right) = \frac{1 + 14}{9} = \frac{15}{9} = \frac{5}{3} \). These values match \(-\frac{-2}{3} = \frac{2}{3} \) and \( \frac{5}{3} \), confirming the correctness of the roots.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). The formula is given by:

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

Here, \( a \), \( b \), and \( c \) are coefficients of the quadratic equation, and \( x \) stands for the variables whose values we need to find.
To use the quadratic formula, a student should first identify these coefficients from the equation. By substituting these values into the formula, one can find the roots of the equation, whether they are real or complex.
The formula provides a straightforward method to solving even complex quadratics, making it an essential component in the mathematics toolkit.

Complex Roots

Complex roots arise when the discriminant in the quadratic formula is negative. The discriminant \( \Delta \) is given by \( b^2 - 4ac \). A negative discriminant means the equation has no real roots, leading us to explore complex numbers.

• Complex numbers are expressed in the form \( a + bi \), where \( i \) is the imaginary unit with \( i^2 = -1 \).
• When dealing with a negative discriminant, the roots can be expressed as \( x = \frac{-b \pm i\sqrt{-\Delta}}{2a} \).

In this exercise, the equation \( 3x^2 - 2x + 5 = 0 \) has a discriminant of \(-56\), indicating complex roots.
By simplifying the expression \( i\sqrt{56} \) to \( 2i\sqrt{14} \) and further solving, the complex roots \( \frac{1 \pm i\sqrt{14}}{3} \) were obtained. These roots do not exist on the real number line but are crucial for certain mathematical applications.

Discriminant

The discriminant is a key part of the quadratic formula and determines the nature of the roots of a quadratic equation. It is calculated as \( \Delta = b^2 - 4ac \).

• If \( \Delta > 0 \), the quadratic equation has two distinct real roots.
• If \( \Delta = 0 \), there is exactly one real root (or a repeated real root).
• If \( \Delta < 0 \), there are two complex roots, as there are no real solutions.

In this exercise, the discriminant was \(-56\), leading to complex roots.
Understanding the discriminant is essential because it quickly tells you what kind of solutions to expect. This can help when deciding how to approach solving the equation—whether to look for real number solutions or to explore complex numbers.

Sum and Product of Roots

The sum and product of roots provide a handy way to check if the solutions obtained from any quadratic solving method are correct. These relationships stem directly from Vieta's formulas.

• The sum of the roots \( \alpha + \beta = -\frac{b}{a} \).
• The product of the roots \( \alpha \beta = \frac{c}{a} \).

After finding the roots using the quadratic formula, use these equations to verify:

• For the equation \( 3x^2 - 2x + 5 = 0 \), the sum of the roots should be \( \frac{2}{3} \).
• The product should be \( \frac{5}{3} \), matching the calculated values of the roots.

These checks help ensure the accuracy of your solution, bringing peace of mind that your calculations are correct. This approach is both practical for solving quadratic problems and for confirming your work in exams or assignments.