Problem 29 - 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. $$-2 n^{2}+3 n+5=0$$

Step-by-Step Solution

Answer

The solutions are \(n = -1\) and \(n = 2.5\).

Step 1: Identify coefficients

The quadratic equation is given in the form \[-2n^2 + 3n + 5 = 0\]The coefficients are: \(a = -2\), \(b = 3\), and \(c = 5\). These values will be used in the quadratic formula.

Step 2: Apply Quadratic Formula

Use the quadratic formula to find the solutions for \(n\).The formula is given by:\[n = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\]Plug in the values \(a = -2\), \(b = 3\), and \(c = 5\) into this formula.

Step 3: Calculate Discriminant

First calculate the discriminant \(b^2 - 4ac\):\[b^2 - 4ac = 3^2 - 4(-2)(5) = 9 + 40 = 49\]The discriminant is 49.

Step 4: Solve for Roots

Now substitute back into the quadratic formula:\[n = \frac{{-3 \pm \sqrt{49}}}{2(-2)}\]Compute the roots:\[n = \frac{{-3 + 7}}{-4} = \frac{4}{-4} = -1\] \[n = \frac{{-3 - 7}}{-4} = \frac{-10}{-4} = 2.5\]The solutions to the quadratic equation are \(n = -1\) and \(n = 2.5\).

Step 5: Check Solution with Sum and Product

For roots \(n_1\) and \(n_2\), the sum \(n_1 + n_2\) should equal \(-\frac{b}{a}\) and the product \(n_1 \times n_2\) should equal \(\frac{c}{a}\).Calculate these values:- Sum: \(-1 + 2.5 = 1.5\)- \(-\frac{b}{a} = -\frac{3}{-2} = 1.5\)- Product: \(-1 \times 2.5 = -2.5\)- \(\frac{c}{a} = \frac{5}{-2} = -2.5\)Both the sum and product match the predictions, confirming the correctness of our solutions.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). This formula is helpful when factoring is difficult or impossible. To use the quadratic formula, you substitute the coefficients \(a\), \(b\), and \(c\) from your equation into the formula:

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

The characters \( \pm \) tell us there can be two possible solutions: one adding the discriminant and one subtracting it. It is designed to give both roots of the quadratic equation. Make sure to carefully substitute each coefficient from the quadratic equation, respecting the signs, to avoid errors.
Applying this formula, as in the example, enables us to tackle any solvable quadratic equation with confidence.

Discriminant

The discriminant, represented by the expression \( b^2 - 4ac \) in the quadratic formula, is crucial in determining the nature and number of solutions to a quadratic equation. Here’s what the discriminant tells us:

• If the discriminant is positive, as in our example where it equaled 49, there are two distinct real roots.
• If the discriminant is zero, there is exactly one real root, indicating a repeated or double root.
• If the discriminant is negative, there are no real roots, and the solutions are complex or imaginary.

Knowing the discriminant before solving provides insight into what kind of solutions to expect, and it helps direct the method for solving the quadratic equation. Always calculate this early to guide your problem-solving approach.

Sum and Product of Roots

The sum and product of the roots of a quadratic equation offer a neat way to verify solutions without detailed calculations. Given a quadratic equation \( ax^2 + bx + c = 0 \), the relationships are:

• Sum of the roots \( (n_1 + n_2) = -\frac{b}{a} \)
• Product of the roots \( (n_1 \times n_2) = \frac{c}{a} \)

In the context of the example, these calculations confirmed the correctness of the roots obtained from the quadratic formula. The sum of \(-1\) and \(2.5\) matches the expected result of \(1.5\), and their product of \(-2.5\) aligns with the anticipated value from \(\frac{5}{-2}\).
This technique serves as an excellent double-check, ensuring that no computational errors were made in earlier steps.