Problem 86 - Algebra and Trigonometry

Question

Solve equation by the method of your choice. $$ 5 x^{2}=6-13 x $$

Step-by-Step Solution

Answer

The solutions to the equation \(5x^2 + 13x - 6 = 0\) are \(x = 0.4\) and \(x = -3\).

Step 1: Rewrite the equation in standard form

Let's move all the terms to one side of the equation, therefore we will perform two operations: add \(13x\) to both sides and subtract \(6\) from both sides: \(5x^2 + 13x - 6 = 0\)

Step 2: Apply the Quadratic Formula

The Quadratic Formula is given by \(x = \frac{-b \pm \sqrt {b^2 -4ac}}{2a}\). For this problem, \(a = 5\), \(b = 13\), and \(c = -6\). Thus, we apply those into the formula, we obtain \(x = \frac{-13 \pm \sqrt {{13}^2 -4*5*(-6)}}{2*5}\).

Step 3: Simplify the roots

Applying the operations present in above formula, we first calculate the discriminant \(\Delta = b^2 -4ac = 169 - (-120) = 289\). Since the discriminant is a perfect square, we can take its square root to obtain \(17\). Thus, the equation simplifies to \(x = \frac{-13 \pm 17}{10}\). This gives us two roots: \(x_1 = \frac{4}{10} = 0.4\) and \(x_2 = \frac{-30}{10} = -3\).

Key Concepts

Quadratic Formula

When you're faced with a quadratic equation like the one in this exercise, the quadratic formula is one of the most powerful tools at your disposal. It allows you to find the roots of any quadratic equation of the form \(ax^2 + bx + c = 0\). Here's the formula in all its glory:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula is convenient because you only need to know the coefficients \(a\), \(b\), and \(c\), which are the numbers in front of \(x^2\), \(x\), and the constant term, respectively. Then, it's just a matter of plugging them in and performing the arithmetic to find the values of \(x\). This is particularly helpful when other methods, like factoring, wouldn't work. It's a universal solution for all quadratic equations!

Discriminant

The discriminant is like a special detective in mathematics that helps us understand the nature of the roots of a quadratic equation. It's the part of the quadratic formula under the square root sign, \(b^2 - 4ac\). You'll find that the discriminant can tell you:

• If it's positive, the quadratic equation has two distinct real roots.
• If it's zero, there is exactly one real root, also known as a repeated or double root.
• If it's negative, the roots are not real numbers but complex ones.

In our exercise, the discriminant is \(289\), a positive perfect square. This means we have two real and distinct roots. Understanding the discriminant can guide your expectations of the solutions you'll find.

Roots of Equations

The roots of a quadratic equation are the values of \(x\) that make the equation true; effectively, they are where the curve represented by the equation crosses the x-axis. For our exercise, after applying the quadratic formula, we determined two roots: \(x_1 = 0.4\) and \(x_2 = -3\). These roots mean that when you substitute \(x_1\) and \(x_2\) back into the original equation, the equality holds true. Real roots are the points where the parabola intersects the x-axis, providing valuable insights into the behavior of the equation graphically. Knowing how to find these roots through the quadratic formula is a vital skill, offering solutions even when other methods, like graphing, aren't feasible. Understanding each root's implication helps in various applications of quadratic equations, from physics to engineering.