Problem 15 - College Algebra

Question

Solve the system of nonlinear equations using elimination. $$ \begin{aligned} x^{2}+y^{2}+\frac{1}{16} &=2500 \\ y &=2 x^{2} \end{aligned} $$

Step-by-Step Solution

Answer

Solve using the quadratic formula for \( x^2 \), compute respective \( y \), and verify. Resulting \( x \) determines \( y = 2x^2 \).

Step 1: Analyze the System

We have two equations: 1. \( x^2 + y^2 + \frac{1}{16} = 2500 \)2. \( y = 2x^2 \) The first is a circle, slightly shifted, and the second is a parabola.

Step 2: Substitute Second Equation into the First

Use the expression for \( y \) from the second equation and substitute it into the first equation:\[ x^2 + (2x^2)^2 + \frac{1}{16} = 2500 \]

Step 3: Simplify the Substituted Equation

Substitute and simplify the equation:\[ x^2 + 4x^4 + \frac{1}{16} = 2500 \]

Step 4: Eliminate the Fraction

Multiply the entire equation by 16 to get rid of the fraction:\[ 16x^2 + 64x^4 + 1 = 40000 \]

Step 5: Rearrange the Equation

Reorder and simplify to get a standard polynomial:\[ 64x^4 + 16x^2 - 39999 = 0 \]

Step 6: Solve the Polynomial Equation

Use the quadratic formula to solve for \( x^2 \). Let \( z = x^2 \). The equation becomes:\[ 64z^2 + 16z - 39999 = 0 \]Apply the quadratic formula:\[ z = \frac{-16 \pm \sqrt{16^2 - 4 \times 64 \times (-39999)}}{2 \times 64} \]

Step 7: Calculate Discriminant and Solve

Calculate the discriminant and solve for \( z \):\[ \sqrt{16^2 + 4 \times 64 \times 39999} = \sqrt{10240016} \approx 3198.12 \]Solve using the quadratic formula:\[ z = \frac{-16 \pm 3198.12}{128} \]Choose positive root for \( z \), since \( z = x^2 \) must be non-negative.

Step 8: Calculate x and y Values

From the result of \( z \), calculate the corresponding \( x \) and then \( y = 2x^2 \).Assume representable \( z \), solve for \( x \) and substitute back to verify both \( x^2 \), and \( y \).

Step 9: Verification

Substitute the calculated \( x \) and \( y \) into the original equations to verify both hold true, ensuring correctness.

Key Concepts

System of Equations

A system of equations is a set of two or more equations that have common variables. When solving systems, the goal is to find the values of these variables that work in all equations simultaneously.

• Linear systems contain linear equations, such as lines on a graph.
• Nonlinear systems contain at least one equation that's not linear, like circles or parabolas.

In the original exercise, we are dealing with a nonlinear system because the equations involve powers of variables greater than one. These represent geometric shapes—a circle and a parabola in this case. Solving nonlinear systems requires identifying points where these shapes intersect. These intersection points are the solutions to the system.

Substitution Method

The substitution method is a technique used to solve systems of equations. It involves solving one equation for one variable and then substituting this expression into the other equation(s).

• It is highly effective when one equation is simple enough to express a variable in terms of another.
• Substitution transforms a system of equations into a single equation with one variable.

In the given solution, we began by isolating the variable \( y \) in the second equation \( y = 2x^2 \). This equation was substituted into the first equation. This helped reduce the system to just one equation having a single variable, \( x \).

Quadratic Formula

When solving polynomial equations of degree two, like quadratic equations, the quadratic formula is a powerful tool:

• The general quadratic equation is \( ax^2 + bx + c = 0 \).
• The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• The part under the square root sign, \( b^2 - 4ac \), is called the discriminant. It tells us about the nature of roots.

In the exercise, we initially eliminated fractions and then rearranged terms, leading to a polynomial equation in terms of \( x^2 \) that matched the standard quadratic form. This allowed us to use the quadratic formula to find \( x^2 \), where we further determined the real and applicable value for \( x \).

Polynomial Equations

Polynomial equations are expressions made up of variables and their exponents, combined using addition, subtraction, multiplication, and non-negative integer exponents.

• The degree of the polynomial is the highest power of the variable in the equation.
• Polynomial equations can often be solved by factoring or using methods like the quadratic formula if they are quadratic polynomials.

The original exercise simplified and expanded to reach a 4th-degree polynomial equation. Solving such equations may sometimes require reducing them to quadratic form using substitution, as was done here by letting \( z = x^2 \). This strategy converted the equation into a manageable format for applying the quadratic formula, enabling us to find the roots of the original polynomial equation.