Problem 55 - Beginning and Intermediate Algebra

Question

Find the discriminant for each quadratic equation. Use it to tell whether the equation can be solved using the zero-factor property, or the quadratic formula should be used instead. Then solve each equation. (a) \(3 x^{2}+13 x=-12\) (b) \(2 x^{2}+19=14 x\)

Step-by-Step Solution

Answer

For equation (a): Solutions are \(-\frac{4}{3}\) and \(-3\). For equation (b): Solutions are \(\frac{7 + \sqrt{11}}{2}\) and \(\frac{7 - \sqrt{11}}{2}\).

Step 1: Introduction to the Discriminant

The discriminant of a quadratic equation in the form \(ax^{2} + bx + c = 0\) is given by \(D = b^{2} - 4ac\). Depending on the value of the discriminant, we can determine the nature of the roots and how to solve the equation.

Step 2: Convert Equations to Standard Form

To use the discriminant formula, first convert each equation to the standard form \(ax^{2} + bx + c = 0\). (a) For the equation \(3x^{2} + 13x = -12\), add 12 to both sides: \(3x^{2} + 13x + 12 = 0\). (b) For the equation \(2x^{2} + 19 = 14x\), subtract 14x from both sides: \(2x^{2} - 14x + 19 = 0\).

Step 3: Calculate the Discriminant for Each Equation

(a) For \(3x^{2} + 13x + 12 = 0\), identify \(a = 3, b = 13, c = 12\). Substitute these values into the discriminant formula: \(D = 13^{2} - 4 \cdot 3 \cdot 12 = 169 - 144 = 25\). (b) For \(2x^{2} - 14x + 19 = 0\), identify \(a = 2, b = -14, c = 19\). Substitute these values into the discriminant formula: \(D = (-14)^{2} - 4 \cdot 2 \cdot 19 = 196 - 152 = 44\).

Step 4: Determine the Method to Solve Each Equation

If the discriminant \(D > 0\), the quadratic equation has two distinct real roots and can be solved using the quadratic formula. Since both \(D = 25\) and \(D = 44\) are greater than zero, use the quadratic formula for both equations: \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\).

Step 5: Solve Equation (a) Using the Quadratic Formula

For \(3x^{2} + 13x + 12 = 0\), plug in \(a = 3, b = 13, c = 12\) into the quadratic formula: \(x = \frac{-13 \pm \sqrt{25}}{6} = \frac{-13 \pm 5}{6}\). This gives two solutions: \(x = \frac{-13 + 5}{6} = \frac{-8}{6} = -\frac{4}{3}\) and \(x = \frac{-13 - 5}{6} = \frac{-18}{6} = -3\).

Step 6: Solve Equation (b) Using the Quadratic Formula

For \(2x^{2} - 14x + 19 = 0\), plug in \(a = 2, b = -14, c = 19\) into the quadratic formula: \(x = \frac{14 \pm \sqrt{44}}{4} = \frac{14 \pm 2\sqrt{11}}{4} = \frac{7 \pm \sqrt{11}}{2}\). This gives two solutions: \(x = \frac{7 + \sqrt{11}}{2}\) and \(x = \frac{7 - \sqrt{11}}{2}\).

Key Concepts

Discriminant Explained

Understanding the discriminant in a quadratic equation is essential. It tells us about the nature of the roots of the equation. Given a standard quadratic equation of the form \(ax^{2} + bx + c = 0\), the discriminant, represented by \(D\), is calculated using the formula \(D = b^{2} - 4ac\).

To break it down further:

• If \(D > 0\), the equation has two distinct real roots.
• If \(D = 0\), the equation has one real root (or a repeated root).
• If \(D < 0\), the equation has no real roots, meaning the roots are complex or imaginary.

In the given exercises, we first convert the equations into the standard form and then calculate the discriminant to determine the nature of the roots and the best method to solve the equation. For instance:

• For the equation \(3x^{2} + 13x + 12 = 0\), we found \(D = 25\)
• For the equation \(2x^{2} - 14x + 19 = 0\), we found \(D = 44\)

Both have discriminants greater than zero, indicating they each have two distinct real roots.

The Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations when the discriminant is non-negative. It is given by:

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

Using this formula, we can solve any quadratic equation, provided we know the coefficients \(a\), \(b\), and \(c\).

• For \(3x^{2} + 13x + 12 = 0\) (where \(a = 3\), \(b = 13\), \(c = 12\)): We substitute these values into the formula:
  \[ x = \frac{-13 \pm \sqrt{25}}{6} \]
  This simplifies to:
  \[ x = \frac{-13 \pm 5}{6} \]
  Thus, the solutions are:
  \[ x = -\frac{4}{3} \text{ and } x = -3 \]
• For \(2x^{2} - 14x + 19 = 0\) (where \(a = 2\), \(b = -14\), \(c = 19\)): We substitute these values into the formula:
  \[ x = \frac{14 \pm \sqrt{44}}{4} \]
  This simplifies to:
  \[ x = \frac{7 \pm \sqrt{11}}{2} \]
  Thus, the solutions are:
  \[ x = \frac{7 + \sqrt{11}}{2} \text{ and } x = \frac{7 - \sqrt{11}}{2} \]

Real Roots of Quadratic Equations

Real roots refer to the solutions of the quadratic equation that are real numbers. The value of the discriminant helps us identify if an equation will have real roots.

For instance, in our exercise, both equations:

• \(3x^{2} + 13x + 12 = 0\)
• \(2x^{2} - 14x + 19 = 0\)

had discriminants greater than zero, indicating both have two distinct real roots. Real roots can be understood as the x-intercepts of the quadratic graph where it crosses the x-axis. This makes the quadratic equation solutions particularly useful in various real-world problems involving projectile motion, area calculations, and more.

By calculating the discriminant and using the quadratic formula, you can determine the exact points where the parabola of the quadratic equation meets the x-axis.