Problem 47 - Intermediate Algebra

Question

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} y^{2}=40-x^{2} \\ y=x^{2}-10 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \((\sqrt{15}, 5), (-\sqrt{15}, 5), (2, -6), (-2, -6)\)."

Step 1: Substitution

Since both equations are expressed in terms of \(y\), substitute the expression for \(y\) from the second equation into the first equation. This gives \((x^2 - 10)^2 = 40 - x^2\).

Step 2: Expand and Simplify

Expand the expression \((x^2-10)^2\) to get \(x^4 - 20x^2 + 100\). Substituting into the equation, we have \(x^4 - 20x^2 + 100 = 40 - x^2\).

Step 3: Rearrange the Equation

Rearrange the terms to form a polynomial equation: \(x^4 - 19x^2 + 60 = 0\).

Step 4: Set as a Quadratic in Form

Let \(u = x^2\). Then the equation becomes \(u^2 - 19u + 60 = 0\). This is a quadratic equation in \(u\).

Step 5: Solving the Quadratic Equation

Use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = -19\), \(c = 60\). This gives \(u = \frac{19 \pm \sqrt{361 - 240}}{2} = \frac{19 \pm \sqrt{121}}{2}\).

Step 6: Determine Values for u

Calculate \(u\) to find \(u = \frac{19 + 11}{2} = 15\) and \(u = \frac{19 - 11}{2} = 4\). Thus, \(x^2 = 15\) or \(x^2 = 4\).

Step 7: Solve for x

Solve \(x^2 = 15\) to get \(x = \pm \sqrt{15}\). Solve \(x^2 = 4\) to get \(x = \pm 2\).

Step 8: Solve for y

Substitute the values of \(x\) back into \(y = x^2 - 10\). For \(x = \pm \sqrt{15}\), \(y = 15 - 10 = 5\). For \(x = \pm 2\), \(y = 4 - 10 = -6\).

Step 9: List All Solutions

The set of solutions for \((x, y)\) is \(\{(\sqrt{15}, 5), (-\sqrt{15}, 5), (2, -6), (-2, -6)\}\).

Key Concepts

Substitution Method

The substitution method is a powerful technique for solving systems of equations, especially when one equation can be easily expressed in terms of another variable. In our given exercise, we had two equations involving the variables \(x\) and \(y\):

• \(y^2 = 40 - x^2\)
• \(y = x^2 - 10\)

This method allows you to "substitute" one equation into the other, eliminating one of the variables. Here, because both equations are expressed in terms of \(y\), we replace \(y\) in the first equation with the expression from the second equation, resulting in a new equation with just one variable.
Instead of dealing separately with two equations, the substitution method condenses the system into a single equation. This usually makes it easier to solve for one of the variables, in this case, \(x\). After we find the values of \(x\), we can substitute back to find corresponding values of \(y\).
The key strength of the substitution method is it reduces complex systems to manageable single-variable equations by rearranging one of the initial equations.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation of the form \( ax^2 + bx + c = 0 \). In this particular exercise, after substitution and simplifying, we obtained a polynomial equation in terms of \(u\), where we set \(u = x^2\). Thus, we have the equation \(u^2 - 19u + 60 = 0\).
Quadratic equations can be solved using several methods: factoring, completing the square, or the quadratic formula. The simplest real-case scenario is when the quadratic factors neatly; otherwise, the quadratic formula provides a powerful tool to solve these equations:

• The quadratic formula is \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In our exercise, this formula helped us find the values for \(u\), giving us the possible values for \(x^2\) once we converted back. This is where the connection between quadratic solving and the problem solution becomes evident, leading us deeper into finding the values of \(x\) and thereby completing the system.

Real Solutions

When solving systems of equations, especially those involving quadratics, finding the "real solutions" refers to those solutions which are actual numbers (as opposed to complex or imaginary numbers). In our exercise, we found solutions for real numbers \(x\) and \(y\) by working through the quadratic equation.
The determinant \(b^2 - 4ac\) in the quadratic formula indicates the nature of the solutions:

• If the determinant is positive, there are two distinct real solutions.
• If it's zero, there's exactly one real solution (a repeated solution).
• If negative, there are no real solutions, only complex numbers.

In the given exercise, solving for \(u\), the positive determinant value \(121\) yielded two distinct solutions for \(u\), returning us to real solutions \(x^2 = 15\) or \(x^2 = 4\). Consequently, these led to real values for \(x\) and their corresponding \(y\) values, showcasing the complete set of solutions for any equations involving real numbers.