Problem 50 - 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 \(x\). $$ 2 x^{2}-x y=3 y^{2}+1 $$

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: Rewrite the Equation in Standard Form

Starting with the given equation:\[ 2x^2 - xy = 3y^2 + 1 \]Rearrange the terms to bring all terms to one side of the equation:\[ 2x^2 - xy - 3y^2 - 1 = 0 \]This is the standard quadratic equation format.

Step 2: Identify Quadratic Coefficients for Part (a)

Identify and label the coefficients for the quadratic term in \(x\):- Coefficient of \(x^2: a = 2\)- Coefficient of \(x: b = -y\)- Constant term: \(c = -(3y^2 + 1)\)

Step 3: Apply the Quadratic Formula for x in Terms of y

The quadratic formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \(a = 2\), \(b = -y\), and \(c = -(3y^2 + 1)\) into the formula:\[ x = \frac{-(-y) \pm \sqrt{(-y)^2 - 4 \cdot 2 \cdot (-(3y^2 + 1))}}{2 \cdot 2} \]Simplify to find:\[ x = \frac{y \pm \sqrt{y^2 + 8(3y^2 + 1)}}{4} \]

Step 4: Simplify the Discriminant

Simplify the expression under the square root:\[ x = \frac{y \pm \sqrt{y^2 + 24y^2 + 8}}{4} \]Combine like terms:\[ x = \frac{y \pm \sqrt{25y^2 + 8}}{4} \]

Step 5: Identify Quadratic Coefficients for Part (b)

Rewrite the original equation, swapping \(x\) and \(y\):\[ 3y^2 + xy - 2x^2 - 1 = 0 \]Identify coefficients for the quadratic in \(y\):- Coefficient of \(y^2: a = 3\)- Coefficient of \(y: b = -x\)- Constant term: \(c = -(2x^2 + 1)\)

Step 6: Apply the Quadratic Formula for y in Terms of x

Use the quadratic formula again for \(y\):\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \(a = 3\), \(b = -x\), and \(c = -(2x^2 + 1)\) into the formula:\[ y = \frac{-(-x) \pm \sqrt{(-x)^2 - 4 \cdot 3 \cdot (-(2x^2 + 1))}}{2 \cdot 3} \]Simplify to find:\[ y = \frac{x \pm \sqrt{x^2 + 24x^2 + 12}}{6} \]

Step 7: Simplify the Discriminant for y in Terms of x

Simplify the expression under the square root:\[ y = \frac{x \pm \sqrt{25x^2 + 12}}{6} \]

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of degree 2. It has the general form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \). Quadratic equations are fundamental in algebra and appear in various scientific and engineering contexts.

Key properties of quadratic equations include:

• They are represented graphically by parabolas.
• The parabola can open upwards or downwards depending on the sign of the coefficient \( a \).
• They may have two, one, or no real roots depending on their discriminant value.

This equation type can always be solved using the quadratic formula, factoring, completing the square, or graphing methods. In our scenario, we specifically turn to the quadratic formula for a structured approach.

Coefficients

Coefficients are numbers or variables that multiply the terms of an equation. In the quadratic equation, these include \( a \), \( b \), and \( c \), which correspond to:

• \( a \) - the coefficient of the \( x^2 \) term, influencing the parabola's width and direction.
• \( b \) - the coefficient of the \( x \) term, affecting the parabola's axis of symmetry.
• \( c \) - the constant term, which impacts the parabola's vertical shift.

Understanding coefficients is crucial because they determine the shape and placement of the quadratic graph.

In our exercise, these coefficients are derived depending on which variable we're expressing, i.e., \( x \) in terms of \( y \), and the reverse. This proficiency allows for swapping variables while maintaining the form and function of the quadratic structure.

Discriminant

The discriminant in a quadratic equation is given by the expression \( b^2 - 4ac \). It reveals the nature and number of solutions or roots the equation possesses. The discriminant is extracted from under the square root in the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Understanding the discriminant helps to predict:

• If \( b^2 - 4ac > 0 \), there are two distinct real roots.
• If \( b^2 - 4ac = 0 \), there is exactly one real root (a repeated root).
• If \( b^2 - 4ac < 0 \), there are no real roots (complex roots instead).

This means the discriminant directly affects the solutions' form. In the exercise, we calculate the discriminant separately for the expressions \( x \) in terms of \( y \) and \( y \) in terms of \( x \), each offering insight into potential solution values.

Solving for Variables

Solving for variables in quadratic equations involves determining the roots or values that satisfy the equation. Using the quadratic formula is one of the most robust methods, as it provides an explicit solution:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( \pm \) implies that two different roots can be derived, considering both the positive and negative solutions.

Steps involve:

• Identifying coefficients \( a \), \( b \), and \( c \).
• Calculating the discriminant \( b^2 - 4ac \).
• Substituting these values into the quadratic formula.

In the exercise, we perform these steps twice: first to express \( x \) in terms of \( y \), and second to express \( y \) in terms of \( x \). Each step carefully respects the respective coefficients and constants, opening a clear pathway to their solutions, no matter how the equation is rearranged or structured.