Problem 120 - Intermediate Algebra

Question

Solve each equation. $$5 x^{2}=6-13 x$$

Step-by-Step Solution

Answer

The solutions are \(x = 0.4\) and \(x = -3\).

Step 1: Write the equation in standard form

First, we need to write the given equation in the standard form of a quadratic equation, which is \(ax^2 + bx + c = 0\). To do this, move all the terms to one side of the equation. Start with the given equation: \[5x^2 = 6 - 13x\]Subtract 6 and add 13x from both sides to obtain:\[5x^2 + 13x - 6 = 0\]

Step 2: Identify the coefficients

Now that the equation is in standard form, identify the coefficients \(a\), \(b\), and \(c\). For the equation \(5x^2 + 13x - 6 = 0\), the coefficients are:- \(a = 5\)- \(b = 13\)- \(c = -6\)

Step 3: Apply the quadratic formula

To solve for \(x\), use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Substitute the values of \(a\), \(b\), and \(c\) into the formula:\[x = \frac{-13 \pm \sqrt{13^2 - 4 \cdot 5 \cdot (-6)}}{2 \cdot 5}\]

Step 4: Calculate the discriminant

Calculate the value inside the square root, known as the discriminant:\[b^2 - 4ac = 13^2 - 4 \cdot 5 \cdot (-6)\]\[= 169 + 120 = 289\]

Step 5: Solve for x using the quadratic formula

Now that we have the discriminant, substitute it back into the quadratic formula:\[x = \frac{-13 \pm \sqrt{289}}{10}\]Since \(\sqrt{289} = 17\), substitute back to find:\[x = \frac{-13 \pm 17}{10}\]

Step 6: Calculate the two possible solutions for x

Finally, solve for \(x\) by calculating the two solutions:1. \(x = \frac{-13 + 17}{10} = \frac{4}{10} = 0.4\)2. \(x = \frac{-13 - 17}{10} = \frac{-30}{10} = -3\)Therefore, the solutions are \(x = 0.4\) and \(x = -3\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. It is used when factoring is complex or impossible. The formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where:

• \(a\), \(b\), and \(c\) are the coefficients from the standard form of the quadratic equation \(ax^2 + bx + c = 0\).
• \(\pm\) indicates that there are typically two solutions.

To use the formula, identify the coefficients and plug them into the formula. Calculate the discriminant (inside the square root) to find the nature of the roots. Finally, solve for \(x\). The solutions from the formula will give you the x-intercepts of the parabola represented by the equation.

Discriminant

The discriminant is the part of the quadratic formula under the square root, \(b^2 - 4ac\). It plays an essential role in understanding the nature of the roots of a quadratic equation.Here's why the discriminant matters:

• If the discriminant (\(b^2 - 4ac\)) is positive, the quadratic equation has two distinct real solutions. Think of these as two separate points where the parabola crosses the x-axis.
• If it is zero, there is exactly one real solution. This means the parabola just touches the x-axis at that point, called a repeated or double root.
• If the discriminant is negative, there are no real solutions; instead, the solutions are complex numbers. In this case, the parabola does not cross the x-axis at all.

Using the example from the original problem, the discriminant \(289\) is positive, indicating two distinct real roots.

Standard Form of Quadratic Equation

To solve a quadratic equation using methods like the quadratic formula, it's crucial to first convert it to the standard form, which is:\[ ax^2 + bx + c = 0 \]In this form:

• \(a\) is the coefficient of \(x^2\).
• \(b\) is the coefficient of \(x\).
• \(c\) is the constant term.

Moving terms to one side of the equation ensures the other side equals zero, allowing you to clearly identify the values of \(a\), \(b\), and \(c\). In the given problem, moving each term correctly transform the equation to \(5x^2 + 13x - 6 = 0\), making it ready for any solving technique, including the quadratic formula. This is an essential step for clarity and correctness in solving quadratic equations.