Problem 50 - 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. $$-2 x^{2}+4 x+1=0$$

Step-by-Step Solution

Answer

Solutions are \( x = 1 \pm \frac{\sqrt{6}}{2} \).

Step 1: Identify Coefficients

The given quadratic equation is \[ -2x^2 + 4x + 1 = 0. \]This can be written in the standard form \[ ax^2 + bx + c = 0, \]where \( a = -2 \), \( b = 4 \), and \( c = 1 \).

Step 2: Write the Quadratic Formula

The quadratic formula to solve \( ax^2 + bx + c = 0 \) is given by \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]

Step 3: Substitute Coefficients into the Formula

Substitute \( a = -2 \), \( b = 4 \), and \( c = 1 \) into the quadratic formula:\[ x = \frac{-4 \pm \sqrt{4^2 - 4(-2)(1)}}{2(-2)}. \]

Step 4: Calculate the Discriminant

Calculate the discriminant, \( b^2 - 4ac \):\[ 4^2 - 4(-2)(1) = 16 + 8 = 24. \]

Step 5: Simplify the Square Root and Solve

Substitute the discriminant back into the equation:\[ x = \frac{-4 \pm \sqrt{24}}{-4}. \]Simplify:\[ \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}. \]Thus,\[ x = \frac{-4 \pm 2\sqrt{6}}{-4}. \]This gives two solutions:1. \[ x = \frac{-4 + 2\sqrt{6}}{-4} = 1 - \frac{\sqrt{6}}{2}, \]2. \[ x = \frac{-4 - 2\sqrt{6}}{-4} = 1 + \frac{\sqrt{6}}{2}. \]

Step 6: Verify Solutions with Sum and Product

The sum of roots for \( ax^2 + bx + c = 0 \) is given by \[ -\frac{b}{a} = \frac{-4}{-2} = 2. \]Check the calculated sum:\[ \left(1 - \frac{\sqrt{6}}{2}\right) + \left(1 + \frac{\sqrt{6}}{2}\right) = 2. \]The product of roots is given by \[ \frac{c}{a} = \frac{1}{-2} = -\frac{1}{2}. \]Check the calculated product:\[ \left(1 - \frac{\sqrt{6}}{2}\right) \left(1 + \frac{\sqrt{6}}{2}\right) = 1 - \left(\frac{6}{4}\right) = -\frac{1}{2}. \]Both values match, verifying the solutions.

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool for solving equations in the form of \( ax^2 + bx + c = 0 \). It's particularly useful when you cannot factor the quadratic equation easily. The formula is presented as:

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

Here, \( b \) and \( c \) are the coefficients and constant of the equation, while \( x \) represents the solutions.

The "\( \pm \)" sign indicates that there are typically two solutions for \( x \). This means, if you plug in your values of \( a \), \( b \), and \( c \) into the formula, you will end up with two potential roots for your equation.

When using the quadratic formula, it’s crucial to perform each calculation step-by-step to avoid errors, especially when simplifying the square root part of the equation.

Discriminant

The discriminant is the part of the quadratic formula under the square root:

• \( b^2 - 4ac \)

The discriminant provides valuable information about the nature of the roots of a quadratic equation.

Here's how to interpret it:

• If \( b^2 - 4ac > 0 \), the quadratic equation has two distinct real roots.
• If \( b^2 - 4ac = 0 \), there is one real root or a double root.
• If \( b^2 - 4ac < 0 \), the equation has no real roots; the roots are complex or imaginary.

In our specific exercise, we calculated \( b^2 - 4ac = 24 \), which is greater than zero, indicating two distinct real solutions. This guides us in knowing that there will be two viable solutions when we use the quadratic formula.

Sum and Product of Roots

The sum and product of roots offer a quick check to verify your solutions for a quadratic equation. These concepts derive from Vieta's formulas. For a quadratic equation \( ax^2 + bx + c = 0 \):

• The sum of the roots \( (r_1 + r_2) \) is \( -\frac{b}{a} \).
• The product of the roots \( (r_1 \cdot r_2) \) is \( \frac{c}{a} \).

These calculations can confirm the accuracy of the roots obtained from the quadratic formula.

In the original exercise, the sum of roots was calculated as \( 2 \), and the product as \( -\frac{1}{2} \). Both outcomes matched the results from our formula solutions:

• The sum: \( \left(1 - \frac{\sqrt{6}}{2}\right) + \left(1 + \frac{\sqrt{6}}{2}\right) = 2 \).
• The product: \( \left(1 - \frac{\sqrt{6}}{2}\right) \cdot \left(1 + \frac{\sqrt{6}}{2}\right) = -\frac{1}{2} \).

Solving Quadratic Equations

Solving quadratic equations can be approached in several ways, including factoring, completing the square, and using the quadratic formula. The quadratic formula, however, is the most universal method.

To efficiently solve a quadratic equation using the quadratic formula, follow these steps:

• Identify the coefficients \( a \), \( b \), and \( c \) from the equation \( ax^2 + bx + c = 0 \).
• Compute the discriminant \( b^2 - 4ac \) to understand the nature of the roots.
• Utilize the quadratic formula to find the solutions.
• Simplify the results, especially the square root part, to ensure accurate solutions.

For this exercise, the equation \( -2x^2 + 4x + 1 = 0 \) was solved using these methods, yielding two distinct solutions: \( x = 1 - \frac{\sqrt{6}}{2} \) and \( x = 1 + \frac{\sqrt{6}}{2} \). By verifying these with the sum and product of roots, you confirm the solutions are accurate. This comprehensive approach ensures no step is overlooked in reaching the correct answer.