Problem 49 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 49-50: Use the quadratic formula to solve the equation for (a) \(x\) in terms of \(y\) and (b) \(y\) in terms of \(x\). $$ 4 x^{2}-4 x y+1-y^{2}=0 $$

Step-by-Step Solution

Answer

For (a) \(x = \frac{y \pm \sqrt{2y^2-1}}{2}\); for (b) \(y = -2x \mp 2\sqrt{5x^2+1}\).

Step 1: Identify the coefficients when solving for x

The given equation is \(4x^2 - 4xy + 1 - y^2 = 0\). When solving for \(x\), it can be rewritten in the standard quadratic form \(ax^2 + bx + c = 0\) with: \(a = 4\), \(b = -4y\), and \(c = 1 - y^2\).

Step 2: Apply the quadratic formula for x

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the identified coefficients: \[x = \frac{-(-4y) \pm \sqrt{(-4y)^2 - 4 \cdot 4 \cdot (1-y^2)}}{2 \cdot 4}\]Simplify to give the solutions for \(x\).

Step 3: Simplify the expression for x

Calculate the discriminant and simplify:\((-4y)^2 = 16y^2\) and \(-4 \cdot 4 \cdot (1-y^2) = -16(1-y^2)\).Thus, the discriminant \(\Delta = 16y^2 - 16 + 16y^2 = 32y^2 - 16\).Now solve:\[x = \frac{4y \pm \sqrt{32y^2 - 16}}{8}\]This gives:\[x = \frac{4y \pm 4\sqrt{2y^2 - 1}}{8}\]Further simplifying:\[x = \frac{y \pm \sqrt{2y^2 - 1}}{2}\].

Step 4: Identify the coefficients when solving for y

Rearrange the equation to solve for \(y\) in terms of \(x\): \(-y^2 - 4xy + 4x^2 + 1 = 0\).Recognizing this again as a quadratic equation in\(y\), the coefficients are: \(a = -1\), \(b = -4x\), and \(c = 4x^2 + 1\).

Step 5: Apply the quadratic formula for y

Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), substitute these new coefficients:\[y = \frac{-(-4x) \pm \sqrt{(-4x)^2 - 4(-1)(4x^2 + 1)}}{2(-1)}\]Simplify to solve for \(y\).

Step 6: Simplify the expression for y

Compute the discriminant: \((-4x)^2 = 16x^2\) and \(+16(4x^2 + 1) = 64x^2 + 16\).So, the discriminant \(\Delta = 16x^2 + 64x^2 + 16 = 80x^2 + 16\).Solve:\[y = \frac{4x \pm \sqrt{80x^2 + 16}}{-2}\]Factor and simplify:\[y = \frac{4x \pm 4\sqrt{5x^2 + 1}}{-2}\]Thus:\[y = -2x \mp 2\sqrt{5x^2 + 1}\].

Key Concepts

Solving Quadratic Equations

Quadratic equations are mathematical expressions where the highest exponent of the variable is a square. They take the form of \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are coefficients. Solving quadratic equations means finding the value(s) of the variable that make the equation true. By using the quadratic formula:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

you can find these values. It's a universal tool that works for most quadratic equations. When applying it, be sure to identify the coefficients correctly from the original equation. Depending on the given conditions, you might solve for \(x\) or another variable like \(y\). This flexibility is essential in multi-variable situations, allowing you to find expressions for one variable in terms of another.

Discriminant

The discriminant is a crucial component of the quadratic formula: \(b^2 - 4ac\). It indicates the nature of the roots of a quadratic equation:

• If the discriminant is positive, you have two distinct real roots.
• If it's zero, there is exactly one real root (a repeated root).
• If it's negative, the roots are complex.

Understanding the discriminant helps you quickly ascertain the root's nature without solving the whole equation. In the context of solving for \(x\) or \(y\), compute the discriminant first. For example, when solving for \(x\) in terms of \(y\), we determined the discriminant as \(32y^2 - 16\). This understanding allows you to anticipate the steps needed next in the equation solving process.

Quadratic Equations in Two Variables

Quadratic equations don't always involve a single variable. Often, they feature two, such as \(x\) and \(y\), which adds complexity. These are expressed like \(ax^2 - 4xy + c = 0\). Solving these equations typically involves isolating one variable using the quadratic formula and expressing it in terms of the other.This approach requires:

• Rearranging the equation to make one variable the subject, focusing on distinguishing coefficients for each variable.
• Applying the quadratic formula to derive the expression for one variable.

This exercise requires reconfiguring the equation twice: once for \(x\) in terms of \(y\), and another for \(y\) in terms of \(x\). Each time, you swap the roles of \(a\), \(b\), and \(c\) for the target variable. Understanding each step simplifies the overall process, providing both a solution and a deeper grasp of how equations interplay.