Problem 30 - College Algebra

Question

Solve each system by the method of your choice. $$\begin{aligned} &x+y^{2}=4\\\ &x^{2}+y^{2}=16 \end{aligned}$$

Step-by-Step Solution

Answer

The solution to this system of equations consists of the points (4,0), and (-3, \pm \sqrt{7}).

Step 1: Substitute the first equation into the second

The first equation can be arranged to \(x = 4 - y^2\). Substitute this into the second equation: \((4 - y^2)^2 + y^2 = 16\)

Step 2: Simplify the equation

Expand the brackets and combine similar terms to simplify this equation: \[ 16 - 8y^2 + y^4 + y^2 = 16 \ y^4 -7y^2 = 0 \]

Step 3: Solve for y

Factoring out \(y^2\) leaves \(y^2(y^2 - 7) = 0\). The solutions to this equation are \(y = 0\) and \(y = \pm \sqrt{7}\).

Step 4: Finding the x-values that go with y

Substitute the y-values into the first equation to find the corresponding x-values. For \(y = 0\), we get \(x = 4\). For \(y = \pm \sqrt{7}\), we get \(x = 4 - 7 = -3\). So, our solutions are (4,0) and (-3, \pm \sqrt{7}).

Key Concepts

Substitution Method

The substitution method is a technique used to solve systems of equations by manipulating one equation to express one variable in terms of another. This allows the substitution of this expression into another equation, reducing the number of variables and simplifying the system. In our exercise, the substitution method was used strategically with the first equation. Here's how it works step-by-step:

• First, solve one of the equations for one variable. In the given problem, the equation \(x + y^2 = 4\) was manipulated to express \(x\) in terms of \(y\): \(x = 4 - y^2\).
• Next, substitute this expression for \(x\) into the second equation: \((4 - y^2)^2 + y^2 = 16\). This substitution produces an equation with only one variable, which is easier to solve.

This method significantly reduces complexity, as it turns a two-variable problem into a single-variable problem. It's very useful for systems that can be rearranged to isolate variables easily. Always check which variable is easier to isolate before proceeding with substitution.

Quadratic Equations

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). In our exercise, we encounter a quadratic form after substitution: \(y^4 - 7y^2 = 0\). Here's how to analyze such equations:
- Although this isn't in the standard quadratic form, note that it can be viewed as a quadratic by substituting \(z = y^2\), thus turning the equation into \(z^2 - 7z = 0\).- Solving this quadratic equation involves standard methods like factoring, using the quadratic formula, or completing the square when applicable. In this case, factoring is the most straightforward approach.Quadratics can often appear disguised in higher-degree polynomial equations, as seen in this exercise. Recognizing them can simplify solutions considerably. Always look for opportunities to simplify expressions into a recognizable form.

Factoring

Factoring is a mathematical process of breaking down an expression into multipliers that, when multiplied together, produce the original expression. This is crucial in solving equations, especially quadratics. In our exercise, factoring helps solve \(y^4 - 7y^2 = 0\). Let's break down the process:

• Identify a common factor in all terms of the polynomial, in this case, \(y^2\).
• Factor out \(y^2\) from each term: \(y^2(y^2 - 7) = 0\).
• Set each factor equal to zero to find solutions: \(y^2 = 0\) gives \(y = 0\), and \(y^2 - 7 = 0\) gives \(y = \pm \sqrt{7}\).

Factoring simplifies equations and leads to possible solutions, especially when dealing with polynomial equations. It’s an efficient way to find roots when applicable, saving time compared to other methods like the quadratic formula. When facing any polynomial, look for common factors as your first step towards simplification.