Problem 52 - Precalculus : Building Concepts and Connections

Question

Solve the quadratic equation by using the quadratic formula. Find only real solutions. $$-\frac{1}{3} x^{2}-3 x+9=0$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x_1 = - 3 \times \frac{3}{2} - \sqrt{21} \times \frac{3}{2}\) and \(x_2 = - 3 \times \frac{3}{2} + \sqrt{21} \times \frac{3}{2}\).

Step 1: Identify the coefficients

In the quadratic equation \(-\frac{1}{3}x^2 -3x +9 =0\), the coefficients are \(a=-\frac{1}{3}\), \(b=-3\), and \(c=9\) respectively.

Step 2: Substitute the coefficients into the quadratic formula

Put the coefficients \(a\), \(b\), and \(c\) in the quadratic formula. The equation becomes: \[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(-\frac{1}{3})(9)}}{2(-\frac{1}{3})} \]. Simplifying further, we get:\[ x = \frac{3 \pm \sqrt{9 - (-12)}}{- \frac{2}{3}} \]

Step 3: Simplify the equation

Now, simplify the equation further:\[ x = \frac{3 \pm \sqrt{21}}{- \frac{2}{3}} \].Multiply by \(-1\) to get rid of the negative sign in the denominator, it becomes:\[ x = - \frac{3 \pm \sqrt{21}}{\frac{2}{3}} \]Following further simplification, the equation becomes:\[ x = - (3 \pm \sqrt{21}) \times \frac{3}{2} \].

Step 4: Find the roots

Evaluate the equation for the plus and minus in the equation. For the plus:\[ x_1 = - 3 \times \frac{3}{2} - \sqrt{21} \times \frac{3}{2} \]For the minus:\[ x_2 = - 3 \times \frac{3}{2} + \sqrt{21} \times \frac{3}{2} \]

Key Concepts

Solving Quadratic Equations

Quadratic equations are mathematical expressions of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are coefficients, and \( a \) is not zero. Solving these equations means finding the value(s) of \( x \) that satisfy the equation. The quadratic formula, which provides a method for solving quadratic equations, is expressed as \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Here is how you can apply the quadratic formula step-by-step:

• First, make sure the equation is in the standard quadratic form.
• Next, identify the coefficients of the quadratic equation, which are usually given the symbols \( a \), \( b \), and \( c \).
• Substitute these coefficients into the quadratic formula.
• Simplify the expression under the radical sign, known as the discriminant, which is \( b^2 - 4ac \).
• Calculate the two potential solutions using the plus and minus signs in the formula.
• Simplify the results to find the real solutions for \( x \).

Finding Real Solutions

Real solutions to a quadratic equation are the actual number values of \( x \) that solve the equation. After simplifying, if the discriminant (the part under the square root in the quadratic formula, \( b^2 - 4ac \)) is positive, there are two different real solutions. If it's zero, there's exactly one real solution, and if it's negative, there are no real solutions, only complex ones.

Using the quadratic formula, you can find the real solutions as follows: Evaluate the expression inside the square root. If it's positive, proceed to solve for \( x \) by considering both the plus and minus versions in the quadratic formula. This will yield two real solutions, which can be the case if the parabola represented by the equation intersects the x-axis in two places.

Identifying Coefficients

In a quadratic equation of the form \( ax^2 + bx + c = 0 \), the coefficients are the numerical factors of each term. The coefficient \( a \) is associated with \( x^2 \), \( b \) with \( x \), and \( c \) is the constant term. Identifying these correctly is crucial as they directly impact the solutions. This process involves looking at the quadratic equation and noting the value in front of each term.

For instance, in the equation \( -\frac{1}{3}x^2 -3x + 9 = 0 \), the coefficient \( a \) is \( -\frac{1}{3} \), the coefficient \( b \) is \( -3 \), and the coefficient \( c \) is \( 9 \). Make sure to include the sign preceding each number to maintain accuracy when substituting these values into the quadratic formula.