Problem 15 - College Algebra with Corequisite Support

Question

For the following exercises, solve the system of nonlinear equations using elimination. $$ \begin{array}{l} x^{2}+y^{2}+\frac{1}{16}=2500 \\ y=2 x^{2} \end{array} $$

Step-by-Step Solution

Answer

Substitute \(y = 2x^2\), simplify to find \(x\), then solve for \(y\). Verify solutions with original equations.

Step 1: Understand the System of Equations

We have two equations: \( x^2 + y^2 + \frac{1}{16} = 2500 \) and \( y = 2x^2 \). The first is a circle equation, and the second is a parabola equation. Our goal is to find points \((x, y)\) where both equations are satisfied.

Step 2: Substitute the Second Equation into the First

Substitute \(y = 2x^2\) into the first equation, yielding \(x^2 + (2x^2)^2 + \frac{1}{16} = 2500\). Simplify this to get a single equation in terms of \(x\).

Step 3: Simplifying the Substituted Equation

Expand and simplify the expression after substitution: \(x^2 + 4x^4 + \frac{1}{16} = 2500\). This simplifies to \(4x^4 + x^2 + \frac{1}{16} = 2500\).

Step 4: Isolate the Polynomial Equation

Subtract \(\frac{1}{16}\) from both sides to isolate the polynomial: \(4x^4 + x^2 = 2500 - \frac{1}{16}\). Convert \(2500\) to a fraction: \(2500 = 40000/16\), and compute the difference.

Step 5: Solve for \(x\)

Simplify the right side: \(40000/16 - 1/16 = 39999/16\). Thus, the equation becomes \(4x^4 + x^2 = 39999/16\). To solve for \(x\), substitute into a numerical or symbolic solver as this is a quartic equation.

Step 6: Solutions Verification

Once \(x\) values are found, substitute them back into \(y = 2x^2\) to find corresponding \(y\) values. Verify solutions by checking if \(x^2 + y^2 + \frac{1}{16} = 2500\) holds true for found \((x, y)\) values.

Key Concepts

System of Equations

In mathematics, a 'system of equations' refers to a set of two or more equations that share the same variables. Solving these systems involves finding the values of the variables that satisfy all equations simultaneously.
Such systems can be categorized into linear and nonlinear equations. Linear systems contain variables raised to the power of one, whereas nonlinear systems, like in our problem, involve variables raised to higher powers, exponential, logarithmic, or other non-linear forms.
The given problem illustrates a nonlinear system, where we have two different types of equations:

• An equation of a circle: \(x^2 + y^2 + \frac{1}{16} = 2500\).
• An equation of a parabola: \(y = 2x^2\).

Finding solutions requires using techniques like substitution or elimination to handle these interconnected equations.

Elimination Method

The elimination method is a fundamental technique for solving systems of equations. It involves manipulating the equations in a way that eliminates one of the variables, allowing us to solve for another.
This method is often used with linear systems, but it can be adapted for nonlinear systems as well, as shown in the given problem.
Here’s how the elimination method works in this context:

• Firstly, the second equation \(y = 2x^2\) is substituted into the first equation.
• This substitution transforms the system into a single equation involving only the variable \(x\).
• The modified equation is then simplified to isolate the terms involving \(x\).

This method helps reduce the complexity of multivariable problems, ultimately making them more manageable to solve analytically or numerically.

Quartic Equations

Quartic equations are polynomial equations of degree four. They take the general form: \[ ax^4 + bx^3 + cx^2 + dx + e = 0 \] where the highest exponent on the variable \(x\) is four. Quartic equations can be more complex to solve compared to quadratic or cubic equations due to their higher degree.
In the provided solution, after substituting the parabola equation into the circle's equation, we arrive at a quartic equation: \[ 4x^4 + x^2 = \frac{39999}{16} \] Solving quartic equations may involve:

• Factoring
• Numerical methods (e.g., using a calculator or computer software)
• Special formulas for certain types

Once we find \(x\), the solution is verified by substituting back into the original equations to ensure the results satisfy both the circle and the parabola equations.