Problem 56 - Intermediate Algebra

Question

Solve each quadratic equation using the method that seems most appropriate. $$ 9 x^{2}+18 x+5=0 $$

Step-by-Step Solution

Answer

The roots are \(x = -\frac{1}{3}\) and \(x = -\frac{5}{3}\).

Step 1: Identify the Standard Form

The given quadratic equation is in the form \(ax^2 + bx + c = 0\), where \(a = 9\), \(b = 18\), and \(c = 5\).

Step 2: Calculate the Discriminant

The discriminant \(D\) is calculated using the formula \(D = b^2 - 4ac\). Plugging in the values, we have \(D = 18^2 - 4(9)(5) = 324 - 180 = 144\).

Step 3: Check the Discriminant

Since the discriminant \(D = 144\) is a perfect square, it is suitable to use the quadratic formula for finding the real roots.

Step 4: Apply the Quadratic Formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{D}}{2a}\). Substituting the values, we have \(x = \frac{-18 \pm \sqrt{144}}{18}\).

Step 5: Simplify the Expression

First, calculate \(\sqrt{144} = 12\). Now substitute this back into the quadratic formula to get \(x = \frac{-18 \pm 12}{18}\). This will give two solutions: \(x = \frac{-18 + 12}{18}\) and \(x = \frac{-18 - 12}{18}\).

Step 6: Solve for Both Roots

For the first root, calculate \(x = \frac{-6}{18} = -\frac{1}{3}\). For the second root, calculate \(x = \frac{-30}{18} = -\frac{5}{3}\).

Step 7: State the Final Roots

The solutions to the quadratic equation \(9x^2 + 18x + 5 = 0\) are \(x = -\frac{1}{3}\) and \(x = -\frac{5}{3}\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool to solve quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). This formula provides a direct method to find the roots, or solutions, of the equation without needing to factorize or complete the square.

The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• \(a\), \(b\), and \(c\) are coefficients from the quadratic equation.
• The term under the square root, \(b^2 - 4ac\), is known as the discriminant.
• The "\(\pm\)" symbol indicates that the quadratic formula provides two potential solutions for \(x\).

To use the quadratic formula, substitute the values of \(a\), \(b\), and \(c\) into it. For example, with the equation \(9x^2 + 18x + 5 = 0\), we substitute \(a = 9\), \(b = 18\), and \(c = 5\) into the formula. This gives us a step-by-step pathway to finding the roots.

Discriminant

The discriminant is a key part of understanding the nature of the roots of a quadratic equation. It is the expression found under the square root in the quadratic formula:\[D = b^2 - 4ac\]The value of the discriminant determines the number and type of roots of the quadratic equation:

• If \(D > 0\), there are two distinct real roots. This means the quadratic equation will intersect the x-axis at two points.
• If \(D = 0\), there is exactly one real root, often referred to as a "repeated" or "double" root. The graph of the equation touches the x-axis at a point.
• If \(D < 0\), there are no real roots, which implies the quadratic equation does not intersect the x-axis, but instead, has two complex roots.

In the example equation \(9x^2 + 18x + 5 = 0\), the discriminant calculated as \(D = 144\) (since \(18^2 - 4 \times 9 \times 5 = 144\)) shows that the equation has two real solutions.

Real Roots

Real roots are solutions to a quadratic equation where the discriminant is positive or zero. These roots are the x-values where the graph of the quadratic equation intersects the x-axis.

• When the discriminant \(D > 0\), the quadratic equation has two distinct real roots.
• When \(D = 0\), the equation has one real root, and the parabola touches the x-axis at a single point.

In our worked example of the equation \(9x^2 + 18x + 5 = 0\):- Calculating using the quadratic formula and substituting the solved discriminant \(\sqrt{144} = 12\), we found the roots to be \(x = -\frac{1}{3}\) and \(x = -\frac{5}{3}\).

These values are the points where the parabola represented by the equation cuts through the x-axis, illustrating two distinct real roots.