Problem 81 - Precalculus Student Solutions Manual 5th

Question

For each equation, ( \(a\) ) solve for \(x\) in terms of \(y,\) and ( \(b\) ) solve for \(y\) in terms of \(x\). $$2 x^{2}+4 x y-3 y^{2}=2$$

Step-by-Step Solution

Answer

For (a): \[x = -y \pm \frac{\sqrt{10y^2 + 4}}{2}\]. For (b): \[y = \frac{2(x \pm \sqrt{16x^2 - 12})}{3}\].

Step 1 Title- Rearrange Original Equation

Start with the given equation: \[2x^2 + 4xy - 3y^2 = 2\] We will solve part (a) first, which requires solving for \(x\) in terms of \(y\).

Step 2 Title- Recognize the Quadratic Form in x

Recognize that the given equation is a quadratic equation in the variable \(x\): \[2x^2 + 4xy - 3y^2 = 2\]

Step 3 Title- Apply the Quadratic Formula

Apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve for \(x\). Here, \(a = 2\), \(b = 4y\), and \(c = -3y^2 - 2\).

Step 4 Title- Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\): \[D = (4y)^2 - 4(2)(-3y^2 - 2) = 16y^2 + 24y^2 + 16 = 40y^2 + 16\]

Step 5 Title- Simplify the Discriminant

Simplify the discriminant: \[D = 40y^2 + 16 = 4(10y^2 + 4) = 4(10y^2 + 4)\]

Step 6 Title- Perform Final Substitution

Substitute back into the quadratic formula and solve for \(x\): \[x = \frac{-4y \pm \sqrt{40y^2 + 16}}{4} = \frac{-4y \pm 2\sqrt{10y^2 + 4}}{4} = \frac{-2y \pm \sqrt{10y^2 + 4}}{2}\] Therefore, \[x = -y \pm \frac{\sqrt{10y^2 + 4}}{2}\]

Step 7 Title- Rearrange for Part (b)

Now, solve part (b) by rearranging the original equation \[2x^2 + 4xy - 3y^2 = 2\] for \(y\) in terms of \(x\).

Step 8 Title- Recognize the Quadratic Form in y

Recognize that the given equation is also a quadratic equation in the variable \(y\): \[-3y^2 + 4xy + 2x^2 - 2 = 0\]

Step 9 Title- Apply the Quadratic Formula

Apply the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve for \(y\). Here \(a = -3\), \(b = 4x\), and \(c = 2x^2 - 2\).

Step 10 Title- Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\): \[D = (4x)^2 - 4(-3)(2x^2 - 2) = 16x^2 + 24(2x^2 - 2) = 16x^2 + 48x^2 - 48 = 64x^2 - 48\]

Step 11 Title- Simplify the Discriminant

Simplify the discriminant: \[D = 64x^2 - 48\]

Step 12 Title- Perform Final Substitution

Substitute back into the quadratic formula and solve for \(y\): \[y = \frac{-4x \pm \sqrt{64x^2 - 48}}{-6} = \frac{-4x \pm 4\sqrt{16x^2 - 12}}{-6}\] Therefore, \[y = \frac{2x \pm 2\sqrt{16x^2 - 12}}{3} = \frac{2(x \pm \sqrt{16x^2 - 12})}{3}\]

Key Concepts

Solving Quadratic Equations

Solving quadratic equations is a crucial skill in algebra. These equations have the standard form of ax^2 + bx + c = 0 Quadratic equations are characterized by the highest power of the variable being squared. To find solutions for these equations, we often utilize the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
The formula provides two solutions because of the \(\pm\) sign in front of the square root term. It's essential to isolate the quadratic form first and identify the coefficients \(a\), \(b\), and \(c\).
For example, in solving the equation \(2x^2 + 4xy - 3y^2 = 2\) for \(x\) in terms of \(y\), we recognize that the equation is quadratic in \(x\): - Here, \(a = 2\), \(b = 4y\), and \(c = -3y^2 - 2\).
After recognizing the quadratic form, we apply the quadratic formula to find the values of \(x\). This involves substituting our specific coefficients into the formula and simplifying the results. This makes it easier to handle even the complex quadratic expressions we encounter in a variety of algebra problems.

Discriminant Calculation

The discriminant is an essential part of solving quadratic equations. It is found inside the quadratic formula and expressed as \(D = b^2 - 4ac\)
The discriminant helps determine the nature of the roots of a quadratic equation: - If \(D > 0\), there are two distinct real roots - If \(D = 0\), there is exactly one real root (a repeated root) - If \(D < 0\), the roots are complex and not real Let's calculate the discriminant for our original problem, where we need to solve for \(x\) in terms of \(y\), with \(a = 2\), \(b = 4y\), and \(c = -3y^2 - 2\): \(D = (4y)^2 - 4(2)(-3y^2 - 2) = 16y^2 + 24y^2 + 16 = 40y^2 + 16\)
Once the discriminant is calculated, we substitute it back into the quadratic formula. This assists in simplifying the equation further to find the values of the variable. Understanding how to calculate and use the discriminant allows a deeper comprehension of quadratic equations.

Rearranging Equations

Rearranging equations entails changing the structure of an equation to isolate a particular variable. This skill is critical for solving equations and for finding solutions in terms of another variable. To master this, follow these general steps: - Identify the variable that needs to be isolated - Apply algebraic operations like addition, subtraction, multiplication, and division to both sides of the equation - Simplify the equation step-by-step until the target variable is isolated in the given problem, we first needed to rearrange the equation \(2x^2 + 4xy - 3y^2 = 2\) to solve for \(x\). Recognizing it as a quadratic equation allowed us to use the quadratic formula. Similarly, solving for \(y\) in terms of \(x\) required isolating \(y\). The new equation in terms of \(y\) was \(-3y^2 + 4xy + 2x^2 - 2 = 0\)
By applying the quadratic formula once more to this rearranged equation, and simplifying accordingly, we could then solve for \(y\). Rearranging equations is vital across math and science as it allows flexibility in managing complex expressions and determining relationships between different variables clearly.