Problem 12 - Beginning and Intermediate Algebra

Question

Use the quadratic formula to solve each equation. (All solutions for these equations are real numbers.) $$ 9 x^{2}-6 x+1=0 $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{1}{3}\).

Step 1: Identify coefficients a, b, and c

For the quadratic equation \(ax^2 + bx + c = 0\), identify the coefficients. Here, \(a = 9\), \(b = -6\), and \(c = 1\).

Step 2: Write down the quadratic formula

The quadratic formula is given by \[ x = \frac{-b \, \text{±} \, ed{sqrt{b^2 - 4ac}}}{2a} \].

Step 3: Substitute the coefficients into the formula

Plug the values of \(a = 9\), \(b = -6\), and \(c = 1\) into the quadratic formula: \[ x = \frac{-(-6) \, \text{±} \, ned{sqrt{(-6)^2 - 4 \, \times \, 9 \, \times \, 1}}}{2 \, \times \, 9} \].

Step 4: Simplify under the square root

First, calculate the discriminant: \((-6)^2 - 4 \, \times \, 9 \, \times \, 1 = 36 - 36 = 0\). Thus, the equation inside the square root is 0.

Step 5: Simplify the expression

Since the discriminant is 0, the quadratic formula simplifies to: \[ x = \frac{6}{18} = \frac{1}{3} \].

Step 6: Write down the solution

The solution to the equation 9x^2 - 6x + 1 = 0 is \(x = \frac{1}{3}\).

Key Concepts

quadratic equations

A quadratic equation is a second-degree polynomial equation of the form \(ax^2 + bx + c = 0\). Here, 'a', 'b', and 'c' are constants, with 'a' not equal to zero. The term 'quadratic' comes from 'quad,' meaning square, because the variable is squared (i.e., the highest degree is 2). These equations describe parabolic shapes when graphed on the Cartesian plane.
It’s essential to know that any quadratic equation can have:

• Two distinct real solutions
• One real solution
• No real solutions (but two complex solutions)

The nature of the solutions is determined by the discriminant. The formula for any quadratic equation can help solve it effectively, ensuring you find all possible solutions confidently.

discriminant

The discriminant is a critical component of the quadratic formula. It’s found under the square root in the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). The discriminant is represented by \(b^2 - 4ac\). This value determines the nature of the roots of the quadratic equation:

• If \(b^2 - 4ac > 0\), there are two distinct real solutions.
• If \(b^2 - 4ac = 0\), there is exactly one real solution.
• If \(b^2 - 4ac < 0\), there are no real solutions (the solutions are complex).

Discriminant analysis is crucial because it tells us about the number and type of solutions without having to solve the entire equation. For example, in the exercise given, the discriminant was calculated as \(0\), indicating that the quadratic equation has exactly one real solution.

solution of quadratic equations

To solve a quadratic equation using the quadratic formula, you follow these steps:

• Identify the coefficients \(a\), \(b\), and \(c\) in the equation \(ax^2 + bx + c = 0\).
• Substitute these values into the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Calculate the discriminant \(b^2 - 4ac\).
• Evaluate the expression under the square root and simplify the results.
• Solve for the values of \(x\).

Let's break this down with our example:
For the equation \(9x^2 - 6x + 1 = 0\):
- Coefficients are: \(a = 9\), \(b = -6\), \(c = 1\).
- Substitute into the quadratic formula: \(x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 9 \times 1}}{2 \times 9}\).
- Calculate the discriminant: \((-6)^2 - 4 \times 9 \times 1 = 36 - 36 = 0\).
- Since the discriminant is 0, the quadratic formula simplifies to: \(x = \frac{6}{18} = \frac{1}{3}\).
Thus, the solution to \(9x^2 - 6x + 1 = 0\) is \(x = \frac{1}{3}\). This method makes it straightforward to find whether there are one or two solutions, or if the solutions are real or complex.