Problem 73 - A Graphical Approach to Precalculus with Limits

Question

Solve each nonlinear system of equations analytically. $$\begin{aligned}x^{2}+y^{2} &=10 \\\\-x^{2}+y &=-4\end{aligned}$$

Step-by-Step Solution

Answer

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

Step 1: Identify the equations

The given system of equations consists of: 1. \( x^2 + y^2 = 10 \) 2. \( -x^2 + y = -4 \).

Step 2: Solve for y in the second equation

Rearrange the second equation to express \( y \) in terms of \( x \). Given \( -x^2 + y = -4 \), add \( x^2 \) to both sides to get\( y = x^2 - 4 \).

Step 3: Substitute y in the first equation

Substitute the expression for \( y \) from Step 2 into the first equation. \( x^2 + (x^2 - 4)^2 = 10 \) becomes\( x^2 + (x^4 - 8x^2 + 16) = 10 \).

Step 4: Simplify the equation

Expand and combine like terms in the equation from Step 3:\( x^4 - 7x^2 + 16 = 10 \).Subtract 10 from both sides to get\( x^4 - 7x^2 + 6 = 0 \).

Step 5: Use substitution to solve the equation

Let \( u = x^2 \), then \( x^4 = u^2 \).The equation becomes \( u^2 - 7u + 6 = 0 \).

Step 6: Factor the quadratic equation

Factor the equation \( u^2 - 7u + 6 = 0 \) into:\( (u - 6)(u - 1) = 0 \).Thus, \( u = 6 \) or \( u = 1 \).

Step 7: Solve for x

Recall that \( u = x^2 \).If \( u = 6 \), then \( x^2 = 6 \), giving solutions \( x = \pm \sqrt{6} \).If \( u = 1 \), then \( x^2 = 1 \), giving solutions \( x = \pm 1 \).

Step 8: Solve for y using x values

Using \( y = x^2 - 4 \), substitute the \( x \) values:- For \( x = \sqrt{6} \), \( y = 6 - 4 = 2 \).- For \( x = -\sqrt{6} \), \( y = 6 - 4 = 2 \).- For \( x = 1 \), \( y = 1 - 4 = -3 \).- For \( x = -1 \), \( y = 1 - 4 = -3 \).

Step 9: Write the solutions

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

Key Concepts

Analytical Solution

When you're tasked with solving a nonlinear system of equations analytically, you're exploring a mathematical journey to find exact solutions rather than approximate ones. The key feature of an analytical solution is its ability to express answers in closed forms, like simple numbers or expressions. In our given problem, we deal with two equations:

• A quadratic equation involving both variables where the terms are squared.
• A linear-like equation with one term squared.

Analyzing the problem involves important steps like substitution, simplification, and factoring. Each step is critical in finding the exact intersection points of the two curves represented by the equations. By carrying through with these mathematical operations step-by-step, you uncover specific solutions that satisfy both equations simultaneously.

Substitution Method

The substitution method is a handy technique used to simplify and solve systems of equations by reducing the number of variables. This technique is particularly useful in nonlinear systems. Our goal is to take one variable from any given equation and express it in terms of the other.
For example, in the second equation, \(-x^2 + y = -4\), you can solve for \(y\) by rearranging terms, resulting in \(y = x^2 - 4\). With this expression in hand, you substitute it back into the first equation, \(x^2 + y^2 = 10\).
This crucial step allows you to replace the variable \(y\) with \(x^2 - 4\), transforming the original system into a single equation in terms of \(x\). It's a powerful way to reduce one original equation into something much simpler while ensuring you stay on track towards analytical resolution.

Quadratic Equation

A quadratic equation is often a key part of solving nonlinear systems, especially when dealing with equations involving squared variables. Such an equation typically appears in the form \(ax^2 + bx + c = 0\).
In this exercise, through substitution, the original nonlinear system gets simplified into a quadratic form: \(u^2 - 7u + 6 = 0\). Here,\(u\) represents \(x^2\), allowing us to address the equation through established methods for quadratics.
Recognizing the structure of quadratic equations enables us to solve using various methods, such as factoring or the quadratic formula, to find the roots or solutions for the variable. These roots then provide the necessary information to identify the values of the original variable \(x\). Another feature of quadratic equations is that they can have up to two solutions, a factor that reflects the existence of more than one point of intersection between the graphs of the two original equations.

Factoring

Factoring is a mathematical process of breaking down an equation into simpler, more manageable pieces called factors. It's a crucial part of solving quadratic equations, like when we navigated the expression \(u^2 - 7u + 6 = 0\). Here, factoring involves rewriting the equation in a product form, \((u - 6)(u - 1) = 0\).
Factoring is incredibly useful because it directly leads to identifying solutions by setting each factor to zero. This gives us the roots of the quadratic equation, such as \(u = 6\) and \(u = 1\) in our problem.
When factoring, you're essentially finding the values that make each part equal to zero since anything multiplied by zero equals zero itself. This approach is both straightforward and efficient, aligning perfectly with the needs of analytical problem solving by giving exact answers for \(u\), which can then be used to trace back and find the corresponding \(x\) values, eventually leading to the full solution set for the system.