Problem 46 - Precalculus: Functions and Graphs

Question

Use the quadratic formula to solve the equation for (a) \(x\) in terms of \(y\) and (b) \(y\) in terms of \(x .\) \(2 x^{2}-x y=3 y^{2}+1\)

Step-by-Step Solution

Answer

(a) \(x = \frac{y \pm \sqrt{25y^2 + 8}}{4}\) (b) \(y = \frac{-x \pm \sqrt{25x^2 - 12}}{6}\).

Step 1: Rearrange the Equation for Part (a)

Start by rearranging the equation \(2x^2 - xy = 3y^2 + 1\) to have all terms on one side: \(2x^2 - xy - 3y^2 - 1 = 0\). This equation is now ready to be solved for \(x\) in terms of \(y\).

Step 2: Identify Coefficients for Quadratic Formula (Part a)

For the given quadratic equation \(2x^2 - xy - 3y^2 - 1 = 0\), identify the coefficients for the quadratic formula \(ax^2 + bx + c = 0\). Here, \(a = 2, b = -y\), and \(c = -(3y^2 + 1)\).

Step 3: Apply Quadratic Formula for (a)

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the coefficients: \(x = \frac{-(-y) \pm \sqrt{(-y)^2 - 4(2)(-(3y^2 + 1))}}{2 \times 2}\). This simplifies to \(x = \frac{y \pm \sqrt{y^2 + 8(3y^2 + 1)}}{4}\).

Step 4: Simplify the Discriminant for (a)

Simplify the expression inside the square root: \(y^2 + 8(3y^2 + 1) = y^2 + 24y^2 + 8\). This gives the expression \(25y^2 + 8\).

Step 5: Find Solution for (a)

Substitute back to find \(x\) in terms of \(y\): \(x = \frac{y \pm \sqrt{25y^2 + 8}}{4}\). These are the solutions for \(x\).

Step 6: Rearrange the Equation for Part (b)

Rearrange the same original equation to express it as a quadratic in terms of \(y\): \(3y^2 + xy + 1 - 2x^2 = 0\). This equation can be solved for \(y\) in terms of \(x\).

Step 7: Identify Coefficients for Quadratic Formula (Part b)

For the quadratic equation \(3y^2 + xy + 1 - 2x^2 = 0\), identify \(a = 3, b = x\), and \(c = 1 - 2x^2\).

Step 8: Apply Quadratic Formula for (b)

Use the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute: \(y = \frac{-x \pm \sqrt{x^2 - 4(3)(1 - 2x^2)}}{6}\).

Step 9: Simplify the Discriminant for (b)

Simplify the square root: \(x^2 - 12 + 24x^2 = 25x^2 - 12\).

Step 10: Find Solution for (b)

Finally, substitute back to find \(y\) in terms of \(x\): \(y = \frac{-x \pm \sqrt{25x^2 - 12}}{6}\). These give the solutions for \(y\).

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra. They typically take the form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable to solve for. Quadratic equations can have:

• Two solutions
• One solution
• No real solutions

The solutions are found using the quadratic formula. Quadratic equations are everywhere. From physics to finance. They help us model real-world situations. In the problem given, we tackled rearranged versions to find \( x \) in terms of \( y \) and vice versa. This requires expressing the equation in a form suitable for the quadratic formula. Thus, it demands careful manipulation and understanding of the equation structure.

Discriminant

The discriminant is a part of the quadratic formula that provides valuable information about the nature of the solutions. It's found using the formula \( b^2 - 4ac \) from the standard form \( ax^2 + bx + c = 0 \). Here’s what the discriminant tells us:

• If it's positive, there are two distinct real solutions.
• If it's zero, there is exactly one real solution.
• If it's negative, there are no real solutions (solutions are complex).

In our exercise, solving for \( x \) in terms of \( y \) resulted in a discriminant of \( 25y^2 + 8 \). Since this expression is always positive, it indicates that there are two real solutions for each value of \( y \). Similarly, when solving for \( y \) in terms of \( x \), the discriminant was \( 25x^2 - 12 \), affecting the number of solutions depending on the value of \( x \). Understanding the discriminant is crucial in predicting the number and type of solutions.

Rearranging Equations

Rearranging equations is a key skill in algebra. It involves moving terms around to form an equation that can be more easily solved. This is done by performing operations that maintain equality, such as:

• Adding or subtracting terms from both sides
• Multiplying or dividing both sides by the same quantity

In our original problem, we rearranged the equation to fit the standard quadratic form. This was necessary to apply the quadratic formula. The first rearrangement was to solve for \( x \) in terms of \( y \): \( 2x^2 - xy - 3y^2 - 1 = 0 \). The process required careful adjustment to isolate terms correctly. Similarly, for finding \( y \) in terms of \( x \), we rearranged to get \( 3y^2 + xy + 1 - 2x^2 = 0 \). Mastery in rearranging equations opens doors to solving a wide range of mathematical problems efficiently.