Problem 28 - Precalculus: Mathematics for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{aligned} x^{2}+2 y^{2} &=2 \\ 2 x^{2}-3 y &=15 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solutions are \((3,1), (-3,1), (\sqrt{6}, \frac{3}{2}), (-\sqrt{6}, \frac{3}{2})\).

Step 1: Solve for y in the second equation

The second equation is \(2x^2 - 3y = 15\). First, solve for \(y\) by isolating it. Add \(3y\) to both sides: \(2x^2 = 3y + 15\). Next, subtract 15 from both sides: \(2x^2 - 15 = 3y\). Finally, divide each term by 3 to solve for \(y\): \(y = \frac{2x^2 - 15}{3}\).

Step 2: Substitute y into the first equation

Now substitute the expression for \(y\) from Step 1 into the first equation \(x^2 + 2y^2 = 2\). This gives \(x^2 + 2\left(\frac{2x^2 - 15}{3}\right)^2 = 2\).

Step 3: Simplify and solve the equation

Simplify the equation from Step 2. Start by expanding the squared term: \(\left(\frac{2x^2 - 15}{3}\right)^2 = \frac{(2x^2 - 15)^2}{9}\). Multiply by 2: \(\frac{2(2x^2 - 15)^2}{9}\). Now the equation is: \(x^2 + \frac{2(2x^2 - 15)^2}{9} = 2\). This results in the polynomial: \(9x^2 + 2(4x^4 - 60x^2 + 225) = 18\). Distribute the 2 and simplify: \(9x^2 + 8x^4 - 120x^2 + 450 = 18\). Combine like terms: \(8x^4 - 111x^2 + 432 = 0\).

Step 4: Factor the polynomial equation

Attempt to factor the polynomial equation \(8x^4 - 111x^2 + 432 = 0\). Use substitution: let \(u = x^2\), so the equation becomes \(8u^2 - 111u + 432 = 0\). Apply the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 8, b = -111, c = 432\). Calculate the discriminant and solve for \(u\). This results in \(u = 9\) or \(u = 6\). Thus, \(x^2 = 9\) or \(x^2 = 6\).

Step 5: Find the values of x

Solve \(x^2 = 9\) to get \(x = 3\) or \(x = -3\). Solve \(x^2 = 6\) to get \(x = \sqrt{6}\) or \(x = -\sqrt{6}\).

Step 6: Solve for y using x values

Use the \(y = \frac{2x^2 - 15}{3}\) expression and the found \(x\) values to solve for \(y\). For \(x = 3\), \(y = 1\); for \(x = -3\), \(y = 1\). For \(x = \sqrt{6}\), \(y = \frac{3}{2}\); for \(x = -\sqrt{6}\), \(y = \frac{3}{2}\).

Step 7: List all solutions

The solutions to the system of equations are the pairs \((x, y)\): \((3, 1)\), \((-3, 1)\), \((\sqrt{6}, \frac{3}{2})\), and \((-\sqrt{6}, \frac{3}{2})\).

Key Concepts

Solving Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(a, b,\) and \(c\) are constants, and \(x\) is the variable you want to solve for. These equations are called quadratics because the variable is squared (the maximum exponent is two). To solve these equations, several techniques can be used depending on the equation's complexity and setup.

The main strategies include:

• Factoring: Expressing the quadratic as a product of binomials.
• Using the quadratic formula: Applying the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Completing the square: Rewriting the equation to highlight a perfect square trinomial.

Each method has its own advantage depending on the coefficients involved in the quadratic equation. Understanding how to use these methods efficiently will make solving quadratic equations much simpler.

Substitution Method

The substitution method is often used in systems of equations, a method where you solve one of the equations for one variable, and then substitute that expression into the other equation. This technique is particularly useful when one equation is easier to manipulate.

By substituting, you transform the system of equations into a single equation with one variable. In our exercise, we transformed the system of equations by solving for \(y\) in one equation and substituting its value into the other equation. This narrowed our focus to a single variable, \(x\), making it simpler to solve. The substitution method is a powerful tool because it systematically reduces complexity by eliminating variables one at a time.

Factoring Polynomials

Factoring polynomials is the process of breaking down a polynomial into simpler polynomials that when multiplied together give the original polynomial. This technique is essential when solving polynomial equations because it allows you to find solutions more easily.

In the exercise solution, factoring was crucial after simplifying the polynomial equation derived from substitution. The polynomial \(8x^4 - 111x^2 + 432 = 0\) needed to be simplified to easily identify solutions. Using a substitution of \(u = x^2\) helped convert the equation into a more common quadratic form, \(8u^2 - 111u + 432 = 0\), which could then be tackled using traditional quadratic solving methods. Factoring, where possible, is often the quickest path to finding the roots of a polynomial equation.

Quadratic Formula

The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is a universal method for finding the roots of any quadratic equation \(ax^2 + bx + c = 0\). This formula comes in handy especially when the quadratic equation cannot be factored easily.

In our exercise, after substituting variables, the formula was utilized in solving the equation \(8u^2 - 111u + 432 = 0\) for \(u\). This led to solutions for \(u\) that we reverted back to solutions for \(x\) since \(u = x^2\). The quadratic formula provides a fail-safe option because it works for every quadratic equation, even when factoring seems too difficult or impossible, making it an invaluable method in solving quadratic equations.