Problem 33 - Algebra 1

Question

Represent the solution graphically. Check the solution algebraically. $$ x^{2}-x=6 $$

Step-by-Step Solution

Answer

The solutions of the quadratic equation \(x^{2}-x-6 = 0\) are \(x= -2\) and \(x =3\). These are verified by substituting back into the original equation and through graphical representation.

Step 1: Rearrange the equation

Rearrange the equation to have all terms on one side: \( x^{2}-x-6 = 0 \)

Step 2: Solve using the quadratic formula

Use the quadratic formula which is given by \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\). In this case \(a=1\), \(b=-1\), and \(c=-6\). So, \(x = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4*1*-6}}{2*1}\) which simplifies to \(x = \frac{1 \pm \sqrt{1 + 24}}{2}\) and gives two solutions \(x = 3\) and \(x = -2\).

Step 3: Check solutions

Check the solutions by substituting back into the original equation. For \(x=3\), substition gives \(3^{2} - 3 = 6\). For \(x=-2\), substitution gives \((-2)^{2}-(-2)=6\). Both solutions verify.

Step 4: Graph the quadratic function

On a graph, plot the function \(f(x) = x^{2}-x-6\). The x-intercepts are \(x= -2\) and \(x =3\), which confirms the algebraic solutions.

Key Concepts

Graphing Solutions

Understanding how to graph solutions to quadratic equations is an essential skill. When graphing, the goal is to visually represent the solutions or roots of the equation. For the equation \( x^2 - x - 6 = 0 \), this involves plotting the related quadratic function \( f(x) = x^2 - x - 6 \) on a coordinate plane. The graph of this function is a parabola.

To visualize this parabola:

• Identify the function's x-intercepts. These are the values of \( x \) that make \( f(x) = 0 \), which have been calculated as \( x = 3 \) and \( x = -2 \).
• The vertex of a parabola in the form \( ax^2 + bx + c \) can be found using \( x = -\frac{b}{2a} \). For this equation, the vertex is at \( x = \frac{1}{2} \). Substitute \( x = \frac{1}{2} \) back into \( f(x) \) to find the corresponding \( y \)-coordinate.
• Plot the points and sketch the parabola. The curve should pass through the x-intercepts and bend over the vertex.

This graph provides a visual confirmation of the calculated algebraic solutions, as the x-intercepts match the real roots found using the quadratic formula.

Quadratic Formula

The quadratic formula is a powerful tool for finding the solutions of quadratic equations. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \). The quadratic formula for finding the roots is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]

In our example, the quadratic equation \( x^2 - x - 6 = 0 \) identifies the coefficients as \( a = 1 \), \( b = -1 \), and \( c = -6 \). By substituting these values into the quadratic formula, we find:

• Calculate the discriminant \( b^2 - 4ac \): \((-1)^2 - 4\times1\times(-6) = 25\).
• The square root of the discriminant \( \sqrt{25} \) is 5.
• Use the quadratic formula to find the roots: \( x = \frac{1 \pm 5}{2} \).
• This results in two solutions: \( x = 3 \) and \( x = -2 \).

The solutions obtained are the points where the graph of the function crosses the x-axis.

Checking Solutions

Confirming solutions algebraically ensures their accuracy and provides a good practice check. Once solutions are derived, they should be substituted back into the original equation to verify correctness. Let’s check our solutions for the equation \( x^2 - x = 6 \):

• Substitute \( x = 3 \): \( 3^2 - 3 = 6 \). Simplifying gives \( 9 - 3 = 6 \), confirming the solution is correct.
• Substitute \( x = -2 \): \((-2)^2 - (-2) = 6 \). Simplifying gives \( 4 + 2 = 6 \), again confirming the solution.

Both examples prove that our calculated solutions are indeed accurate. This verification step is key in solidifying one's understanding and ensuring that no errors have been made in the calculations.