Problem 34 - Algebra and Trigonometry

Question

\(33-40=\) Use the graphical method to find all solutions of the system of equations, rounded to two decimal places. $$ \left\\{\begin{array}{l}{y=x^{2}-4 x} \\ {2 x-y=2}\end{array}\right. $$

Step-by-Step Solution

Answer

Solutions are approximately (5.65, 9.30) and (0.35, -1.30).

Step 1: Understand the Equations

We have two equations: \(y = x^2 - 4x\) and \(2x - y = 2\). The goal is to find all \((x, y)\) pairs that satisfy both equations simultaneously.

Step 2: Rewrite the Second Equation

Rewrite the second equation \(2x - y = 2\) in terms of \(y\): this becomes \(y = 2x - 2\). Now we have two equations both expressed in terms of \(y\).

Step 3: Set the Equations Equal

Since both expressions equal \(y\), set the right sides equal to each other: \(x^2 - 4x = 2x - 2\).

Step 4: Rearrange into a Quadratic Equation

Rearrange the equation \(x^2 - 4x = 2x - 2\) to obtain a standard form quadratic equation: \(x^2 - 6x + 2 = 0\).

Step 5: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -6\), and \(c = 2\). Substitute and solve: \(x = \frac{6 \pm \sqrt{36 - 8}}{2}\). This simplifies to \(x = \frac{6 \pm \sqrt{28}}{2}\), which further simplifies to \(x = \frac{6 \pm 2\sqrt{7}}{2}\). So, \(x = 3 \pm \sqrt{7}\).

Step 6: Calculate Corresponding y Values

Substitute both \(x = 3 + \sqrt{7}\) and \(x = 3 - \sqrt{7}\) back into the equation \(y = 2x - 2\) to find \(y\). For \(x = 3 + \sqrt{7}\), \(y = 2(3 + \sqrt{7}) - 2 = 6 + 2\sqrt{7} - 2 = 4 + 2\sqrt{7}\). For \(x = 3 - \sqrt{7}\), \(y = 2(3 - \sqrt{7}) - 2 = 6 - 2\sqrt{7} - 2 = 4 - 2\sqrt{7}\).

Step 7: Round Solutions

Compute approximate values using the decimal approximation for \(\sqrt{7} \approx 2.65\). Thus, \(y = 4 + 2(2.65) = 9.3\) for \(x = 3 + 2.65\) which is approximately 5.65. For \(x = 0.35\), \(y = 4 - 5.3 = -1.3\). So, the solutions are approximately \((5.65, 9.30)\) and \((0.35, -1.30)\).

Key Concepts

System of Equations

A system of equations is a set of two or more equations with the same variables. In this exercise, we are given two equations involving the variables \(x\) and \(y\). The goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. These simultaneous solutions are called the intersection points of the equations when graphed.

There are various methods to solve systems of equations, including substitution, elimination, and graphical methods. When using the graphical method, we graph each equation on a coordinate plane and look for points where the graphs intersect. These points represent solutions to the system.

In this example, we first equate the equations for \(y\), which allows us to find \(x\)-values where the curves intersect. Then, these \(x\)-values are used to find corresponding \(y\)-values.

Quadratic Equation

A quadratic equation is a polynomial equation of degree 2, which generally has the form \(ax^2 + bx + c = 0\). The equation represents a parabola when graphed on the coordinate plane.

In this exercise, by setting the expressions for \(y\) equal, we derived a quadratic equation: \(x^2 - 6x + 2 = 0\). This equation needs to be solved to find the intersection \(x\)-values.

Quadratic equations can have different forms of solutions, depending on the discriminant \(b^2 - 4ac\). For our specific case, the positive discriminant indicated two real and distinct solutions for \(x\). These solutions are crucial as they provide the horizontal coordinates where the graphical representations intersect.

Quadratic Formula

The quadratic formula is a reliable method to solve quadratic equations when factoring is not straightforward. The formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our equation \(x^2 - 6x + 2 = 0\), we identified the coefficients: \(a = 1\), \(b = -6\), and \(c = 2\). Substituting these into the formula, we calculated the roots of the equation. Solving gives us: \(x = 3 \pm \sqrt{7}\).

The \(\pm\) symbol indicates that there are two solutions or roots, representing the points where the parabola and the line from our system of equations intersect on the graph. These roots are then used to determine the corresponding \(y\) values, completing the solution process.

Solutions of Equations

Solutions of equations refer to the values of the variables that satisfy all equations in a system simultaneously. In the context of the graphical method, solutions are the intersection points on the graph where each pair \((x, y)\) satisfies both equations.

In this example, after finding the \(x\)-values using the quadratic formula, we substitute them back into one of the equations to determine the related \(y\) values. Doing so gives us the specific ordered pairs \((x, y)\) as our final solutions.

We round these solutions to two decimal places: \((5.65, 9.30)\) and \((0.35, -1.30)\), providing a practical and easily understandable representation of the graphically identified intersection points. These rounded values are frequently used in applications where approximate solutions are sufficient for problem-solving purposes.