Problem 82 - 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 \(x\). $$5 x^{2}-6 x y+2 y^{2}=1$$

Step-by-Step Solution

Answer

For (a): \[x = \frac{3y \pm \sqrt{5 - y^2}}{5}\]. For (b): \[y = \frac{3x \pm \sqrt{2 - x^2}}{2}\].

Step 1: Rewrite the equation in standard form

Starting with the given equation: \[5x^2 - 6xy + 2y^2 = 1\].

Step 1 (a) - Treat y as a constant and rewrite the quadratic equation

Rewrite the equation in terms of x while treating y as a constant: \[ax^2 + bx + c = 0\], where \[a = 5\], \[b = -6y\], and \[c = 2y^2 - 1\].

Step 2 (a) - Use the quadratic formula to solve for x

Recall that the quadratic formula is given by \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Plugging in the values of a, b, and c: \[ x = \frac{6y \pm \sqrt{(-6y)^2 - 4 \cdot 5 \cdot (2y^2 - 1)}}{2 \cdot 5}\].

Step 3 (a) - Simplify the expression under the square root

Simplify inside the square root: \[ x = \frac{6y \pm \sqrt{36y^2 - 40y^2 + 20}}{10}\]. Simplifying further. \[ x = \frac{6y \pm \sqrt{-4y^2 + 20}}{10}\].

Step 4 (a) - Further simplification

Factor out the common terms to simplify further: \[ x = \frac{6y \pm \sqrt{4(5 - y^2)}}{10}\] Simplify this to get: \[ x = \frac{6y \pm 2\sqrt{5 - y^2}}{10}\] Finally, simplify the fraction: \[ x = \frac{3y \pm \sqrt{5 - y^2}}{5}\]. This gives us the solutions for x in terms of y.

Step 1 (b) - Treat x as a constant and rewrite the quadratic equation

Rewrite the equation in terms of y while treating x as a constant: \[ay^2 + by + c = 0\], where \[a = 2\], \[b = -6x\], and \[c = 5x^2 - 1\].

Step 2 (b) - Use the quadratic formula to solve for y

Recall that the quadratic formula is given by \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Plugging in the values of a, b, and c: \[ y = \frac{6x \pm \sqrt{(-6x)^2 - 4 \cdot 2 \cdot (5x^2 - 1)}}{2 \cdot 2}\].

Step 3 (b) - Simplify the expression under the square root

Simplify inside the square root: \[ y = \frac{6x \pm \sqrt{36x^2 - 40x^2 + 8}}{4}\]. Simplifying further. \[ y = \frac{6x \pm \sqrt{-4x^2 + 8}}{4}\].

Step 4 (b) - Further simplification

Factor out the common terms to simplify further: \[ y = \frac{6x \pm \sqrt{4(2 - x^2)}}{4}\] Simplify this to get:\[ y = \frac{6x \pm 2\sqrt{2 - x^2}}{4}\] Finally, simplify the fraction: \[ y = \frac{3x \pm \sqrt{2 - x^2}}{2}\]. This gives the solutions for y in terms of x.

Key Concepts

Quadratic Formula

The quadratic formula is a key tool in solving quadratic equations.
Let's recall what it is. For any quadratic equation in the form \[ ax^2 + bx + c = 0 \] the quadratic formula helps you find the roots (solutions) for x using: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here,

• a, b, c are the coefficients of the equation
• The term \[b^2 - 4ac \] is called the discriminant
• The \[ \pm \] indicates that there could be two possible values for x

The quadratic formula is very versatile and can be used to solve any standard quadratic equation.
It’s important to correctly substitute the coefficients and to carefully simplify each part of the formula.

Simplifying Expressions

When using the quadratic formula, you often have to simplify complex expressions. This includes:

• Combining like terms
• Factoring
• Simplifying radicals

For example, in our initial problem, we start with the expression under the square root: \[ \sqrt{36y^2 - 40y^2 + 20} \] By simplifying inside the square root, we get: \[ \sqrt{-4y^2 + 20} \] This simplification process involves basic algebra skills, ensuring that each term is correctly combined.
Further simplification might involve factoring, as seen when we take out common factors in the expression: \[ \sqrt{4(5 - y^2)} \]<> Relax, simplify each step, and the overall solution becomes more manageable!

Precalculus Problem-Solving

In precalculus, problem-solving skills involve understanding and manipulating various mathematical expressions. The essential skills typically include:

• Recognizing function types and their properties
• Using algebraic rules for operations on functions
• Applying the quadratic formula correctly
• Interpreting graphical representations

When dealing with an equation such as \[ 5x^2 - 6xy + 2y^2 = 1, \] you must be able to see it as a quadratic equation in two different variables.
Problem-solving in this context includes treating one variable as a constant while solving for the other.
This demonstrates the ability to manipulate and rearrange equations based on their structures.

Equation Manipulation

Manipulating equations involves various techniques to rewrite or solve them.
This includes techniques such as:

• Substituting and isolating specific variables
• Rewriting equations in standard forms
• Factoring and expanding

In our example, we first rewrote the initial equation \[ 5x^2 - 6xy + 2y^2 = 1 \] into a standard quadratic form in terms of x: \[ 5x^2 - 6xy + (2y^2 - 1) = 0 \] by treating y as a constant.
From here, we systematically applied the quadratic formula and manipulated the resulting equations to simplify our final answer.
Similarly, when treating x as a constant, we rearranged and simplified appropriately to solve for y.
These steps show the power of equation manipulation in solving complex problems.