Problem 10 - Algebra for College Students

Question

Solve each quadratic equation using the method that seems most appropriate to you. $$28-x-2 x^{2}=0$$

Step-by-Step Solution

Answer

The solutions are \(x = -4\) and \(x = 3.5\).

Step 1: Rewrite the equation in standard form

The given equation is \(28-x-2x^2=0\). To write it in standard form \(ax^2 + bx + c = 0\), rearrange the terms: \[-2x^2 - x + 28 = 0\].

Step 2: Identify coefficients

The standard form of the quadratic equation is \(ax^2 + bx + c = 0\). Here, \(a = -2\), \(b = -1\), and \(c = 28\).

Step 3: Use the quadratic formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the values of \(a\), \(b\), and \(c\) into the formula: \[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(-2)(28)}}{2(-2)}\].

Step 4: Calculate the discriminant

The discriminant is \(b^2 - 4ac\). Substitute the values: \[(-1)^2 - 4(-2)(28) = 1 + 224 = 225\].

Step 5: Find the square root of the discriminant

The square root of the discriminant \(225\) is \(15\). This means our equation has real and distinct roots.

Step 6: Solve for x using the quadratic formula

Substitute the square root of the discriminant back into the quadratic formula:\[x = \frac{1 \pm 15}{-4}\].

Step 7: Evaluate the expressions for x

Calculate the two possible values for \(x\):1. \(x = \frac{1 + 15}{-4} = \frac{16}{-4} = -4\)2. \(x = \frac{1 - 15}{-4} = \frac{-14}{-4} = 3.5\).

Key Concepts

Quadratic Formula

The quadratic formula is a foolproof method for solving any quadratic equation, which is written in the form \( ax^2 + bx + c = 0 \). This formula is valuable because it works for any type of quadratic equation, whether the roots are real or complex. To use the quadratic formula, we need to simply plug in the coefficients \( a \), \( b \), and \( c \), which are taken directly from the standard form of the equation. The formula is expressed as:

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

Here's a quick breakdown of what happens:

• \(-b\) flips the sign of the coefficient \( b \).
• \(\pm\) implies two operations—adding and subtracting the square root term.
• The \( \sqrt{b^2 - 4ac} \) term, known as the discriminant, determines the nature of the equation's roots.
• The entire expression is divided by \(2a\) to find the solutions for \(x\).

Using this formula, you can always find values of \(x\) that make the quadratic equation true.

Discriminant

The discriminant is a crucial part of the quadratic formula that can tell us much about the roots of the quadratic equation. It is the expression inside the square root, \(b^2 - 4ac\), and its value determines the nature of the roots.

• If the discriminant is positive (greater than zero), the quadratic equation has two real and distinct roots. This means the parabola intersects the x-axis at two points, as we see in our example with the discriminant value of 225.
• If the discriminant is zero, the quadratic equation has exactly one real root, leading to a parabola that just touches the x-axis (at a vertex).
• Lastly, if the discriminant is negative, the quadratic equation has complex or imaginary roots, meaning the parabola doesn't intersect the x-axis at all.

In our solved equation, with a discriminant of 225, we find that the equation has real and distinct roots. This value indicates not only that solutions exist but also helps us find their real and specific values.

Real and Distinct Roots

When a quadratic equation has real and distinct roots, it means that the solutions for \(x\) are two different real numbers. These roots are where the graph of the quadratic function intersects the x-axis. For instance, in our example, after calculating with the quadratic formula, we found two numbers \(x = -4\) and \(x = 3.5\).

• The fact that these roots are distinct (different from each other) comes from the positive value of the discriminant, as we've calculated before.
• Real numbers allow you to trace where exactly these intersections occur on the parabola's graph.

Real and distinct roots are often what you encounter in practical problems, such as those involving physical trajectories or engineering questions, where the exact crossing points are vital. Always using the quadratic formula ensures you don’t miss any potential solutions.