Problem 23 - Algebra 2 and Trigonometry

Question

In \(18-35,\) find each common solution algebraically. Express irrational roots in simplest radical form. $$ \begin{array}{l}{y=x^{2}-4 x+5} \\ {2 y=x+6}\end{array} $$

Step-by-Step Solution

Answer

The solutions are \((4, 5)\) and \((\frac{1}{2}, \frac{13}{4})\).

Step 1: Express y in terms of x

From the first equation, we have \( y = x^2 - 4x + 5 \). This already expresses \( y \) in terms of \( x \).

Step 2: Substitute for y in the second equation

Substitute \( y = x^2 - 4x + 5 \) from the first equation into the second equation \( 2y = x + 6 \). This gives us: \[ 2(x^2 - 4x + 5) = x + 6 \]

Step 3: Simplify the equation

Expand and simplify the equation obtained in Step 2:\[ 2x^2 - 8x + 10 = x + 6 \] Bring terms on one side to set the equation to 0:\[ 2x^2 - 8x + 10 - x - 6 = 0 \]Which simplifies to:\[ 2x^2 - 9x + 4 = 0 \]

Step 4: Use the quadratic formula

The quadratic equation \( 2x^2 - 9x + 4 = 0 \) is in the form \( ax^2 + bx + c = 0 \) where \( a = 2 \), \( b = -9 \), and \( c = 4 \). The quadratic formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Plugging in the values gives us:\[ x = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 2 \cdot 4}}{4} \]\[ x = \frac{9 \pm \sqrt{81 - 32}}{4} \]\[ x = \frac{9 \pm \sqrt{49}}{4} \]

Step 5: Solve for x

Solve the square root and the quadratic equation:\[ \sqrt{49} = 7 \]Thus, we have:\[ x = \frac{9 + 7}{4} \quad \text{or} \quad x = \frac{9 - 7}{4} \]Solving these gives:\[ x = 4 \quad \text{or} \quad x = \frac{1}{2} \]

Step 6: Find the corresponding y-values

Substitute the x-values back into either equation for \( y \). Use \( y = x^2 - 4x + 5 \):For \( x = 4 \):\[ y = 4^2 - 4 \times 4 + 5 = 16 - 16 + 5 = 5 \]For \( x = \frac{1}{2} \):\[ y = \left(\frac{1}{2}\right)^2 - 4 \times \frac{1}{2} + 5 = \frac{1}{4} - 2 + 5 = \frac{1}{4} - \frac{8}{4} + \frac{20}{4} = \frac{13}{4} \]

Step 7: Write the solution as ordered pairs

The solutions are the pairs \((x, y)\) that satisfy both equations. From the previous steps, the solutions are:\[ (4, 5) \quad \text{and} \quad \left( \frac{1}{2}, \frac{13}{4} \right) \].

Key Concepts

Quadratic Equations

Quadratic equations are fundamental to algebra and appear in the form \( ax^2 + bx + c = 0 \). In a quadratic equation, "\(a\)" is the coefficient of \(x^2\), "\(b\)" is the coefficient of \(x\), and "\(c\)" is the constant term. These equations form a parabola when graphed, and their solutions are the x-values where the parabola intersects the x-axis.
To solve quadratic equations, various methods can be used, such as factoring, completing the square, or employing the quadratic formula. The choice of method often depends on the equation itself and how straightforward it is to manipulate.
Understanding this form is important when solving systems of equations that contain a quadratic component, as it helps in identifying which method will be most efficient.

Substitution Method

The substitution method is a widely-used approach to solve systems of equations, especially when dealing with one linear and one non-linear equation. In this method, the aim is to express one variable in terms of the other from one equation and then substitute that expression into the other equation.
For example, if we know \( y = x^2 - 4x + 5 \), we can substitute this expression for \(y\) in the second equation \(2y = x + 6\), simplifying it into a single equation in terms of \(x\).

• This transforms the system into a quadratic equation, which can be addressed using known techniques.
• The substitution method allows us to break down more complicated systems into manageable pieces.

This is essential when tackling problems involving quadratic equations in systems, facilitating our path to the solution.

Quadratic Formula

The quadratic formula is an essential tool for solving any quadratic equation, provided in its standard form \( ax^2 + bx + c = 0 \). The formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Using this formula, you can find the roots (solutions) of the quadratic equation. Here’s how it works:

• \(-b\) represents the opposite of the coefficient of \(x\).
• The discriminant \(b^2 - 4ac\) determines the nature of the roots (real and distinct, real and equal, or complex).
• The term under the square root, \(\sqrt{b^2 - 4ac}\), is critical in identifying root types.

For the equation \(2x^2 - 9x + 4 = 0\), inserting our values into the quadratic formula allows us to calculate the x-values accurately. Remember, this formula always provides answers, even when other methods are not feasible.

Ordered Pairs

Once solutions for \(x\) are found in a system of equations, the next step is to find corresponding \(y\)-values to form ordered pairs \((x, y)\). This step is crucial as it confirms whether the solutions satisfy both equations in the system.
Using the equation \( y = x^2 - 4x + 5 \), substitute each solution for \(x\) to find its matching \(y\)-value. For example:

• When \(x = 4\), substitute into the equation to get \(y = 5\), forming the pair \((4, 5)\).
• When \(x = \frac{1}{2}\), solving gives \(y = \frac{13}{4}\), resulting in the pair \(\left( \frac{1}{2}, \frac{13}{4} \right)\).

These ordered pairs give the points where the original equations intersect, providing the solutions to the system. They verify the solution's accuracy, ensuring it satisfies the conditions outlined in the original problem.