Problem 96 - College Algebra Essentials

Question

Solve each equation in Exercises \(83-108\) by the method of your choice. $$ 9-6 x+x^{2}=0 $$

Step-by-Step Solution

Answer

The solutions of the given quadratic equation \(9-6 x+x^{2}=0\) are \(x = 3\).

Step 1: Rearranging the Equation

Firstly, rewrite the equation in standard quadratic form, which is \(ax^2 + bx + c = 0\). This gives us: \(x^{2} - 6x + 9 = 0\)

Step 2: Applying the Quadratic Formula

Our coefficients in this equation are \(a=1\), \(b=-6\), and \(c=9\). Substituting these values into the quadratic formula, we have: \[x = \frac{-(-6) \pm \sqrt{(-6)^2-4*1*9}}{2*1}\]

Step 3: Calculating the Roots

Now simplify the equation above to find the roots: \[x = \frac{6 \pm \sqrt{36-36}}{2}\] This simplifies to: \[x = \frac{6 \pm 0}{2}\] Therefore the roots are: \[x = \frac{6}{2}\]

Key Concepts

Solving Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). Solving these equations means finding the values of \( x \) that make the equation true. This process can be done using various methods, such as factoring, completing the square, or the quadratic formula.
In this example, the equation is given as \(9 - 6x + x^2 = 0\). We first rearrange it to the standard form: \( x^2 - 6x + 9 = 0 \).
This standard form makes it easier to apply different techniques to solve the equation. It's crucial to rearrange it correctly, ensuring that all terms are properly aligned on one side of the equation, and the equation equals zero. Once in this form, the quadratic formula becomes applicable, providing a straightforward path to the solutions.

Quadratic Formula

The quadratic formula is a universal tool for solving quadratic equations of any type. It is expressed as:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula works by substituting the coefficients \( a \), \( b \), and \( c \) from the quadratic equation \( ax^2 + bx + c = 0 \).
For the equation \( x^2 - 6x + 9 = 0 \), we identify \( a = 1 \), \( b = -6 \), and \( c = 9 \).
Plugging these values into the formula gives:

• \( x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} \)
• \( x = \frac{6 \pm \sqrt{36 - 36}}{2} \)

The formula simplifies calculations by systematically providing the solutions to any quadratic equation, so long as the terms are correctly input. Ensure each part is calculated step-by-step to avoid common errors, especially in the discriminant \( b^2 - 4ac \). This part determines the nature and number of roots that the quadratic equation can have.

Roots of Quadratic Equations

The term "roots" refers to the solutions of the quadratic equation: the values of \( x \) which satisfy the equation \( ax^2 + bx + c = 0 \). These roots can be real or complex numbers, depending on the value of the discriminant \( b^2 - 4ac \).
In our example, after substituting into the quadratic formula, the roots are calculated to be:
\( \sqrt{36 - 36} = \sqrt{0} \), yielding:

• \( x = \frac{6 \pm 0}{2} \)

This results in a double root or a repeated root: \( x = 3 \).
Having a discriminant of zero indicates that there is exactly one real root. This happens when the vertex of the parabola formed by the quadratic function touches the x-axis at one point, meaning it "bounces off" the axis at that root.
Understanding the concept of roots is key to mastering quadratic equations. Roots can tell us not only where the graph intersects the x-axis, but also important features of the graph itself, such as the vertex and direction of opening.