Problem 36 - College Algebra Essentials

Question

In Exercises \(29-42,\) solve each system by the method of your choice. $$ \left\\{\begin{aligned} x^{3}+y &=0 \\ 2 x^{2}-y &=0 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \((0, 0)\) and \((-2, -8)\).

Step 1: Substitute

Substitute the expression for \(y\) from the second equation into the first equation. This results in: \[x^3 + 2x^2 = 0.\]

Step 2: Factor out a common factor

Factor out \(x^2\), this would enable us to find the roots of this equation.This results in: \[x^2(x + 2) = 0.\]

Step 3: Solve the equation

Having factored the quadratic equation, we can now solve for \(x\). Setting each factor equal to zero gives us the roots of the equation:\[x_1 = 0,\]\[x_2 = -2.\]

Step 4: Substitute back to find y

Substitute the solutions we found for \(x\) into any of the original equations to find the corresponding \(y\) values. Substituting in the first original equation gives:\[y_1 = -0^3 = 0,\]\[y_2 = -(-2)^3 = -8.\]

Key Concepts

Method of Substitution

The method of substitution is a fundamental algebraic technique used to solve systems of equations. This method involves replacing one variable with an equivalent expression from another equation within the system. It is particularly useful when one of the equations is already solved for a particular variable.

When using substitution to solve systems of equations, you usually start by isolating a variable in one equation and then replacing that variable in the other equation. This turns a system of equations into a single equation with a single variable, which is much simpler to solve. Once the value of one variable is found, it is substituted back into one of the original equations to find the corresponding value of the other variable.

The method is widely appreciated for its straightforward approach and the ease with which it can handle even seemingly complex systems that include non-linear equations, such as those involving squared or cubed variables.

Factoring Polynomials

Factoring polynomials is an essential process in algebra that simplifies expressions and enables the solving of equations. When you factor a polynomial, you are looking for two or more simpler expressions that, when multiplied together, give back the original polynomial.

Factoring is particularly helpful in solving equations because it transforms them into a product form, where each factor can potentially be equal to zero. This ability to break down polynomials into factors allows us to find the roots, or solutions, of an equation. Common factoring techniques include taking out a greatest common factor, grouping terms, and using special product rules like the difference of squares or the sum and difference of cubes.

Remaining aware of factoring techniques not only helps in calculations but also in comprehending the inherent relationships within polynomials, which can aid in graphing and understanding their behavior.

Finding Roots of Equations

Finding the roots of equations, or solving an equation, refers to the process of determining the values of the variables that make the equation true. The roots are the solutions at which the graph of the equation crosses or touches the x-axis on a coordinate plane.

For polynomial equations, roots can be found by setting the polynomial equal to zero and solving for the variable. There are multiple methods to find the roots, such as factoring, graphing, or using the quadratic formula for quadratic equations. Depending on the degree of the polynomial, there can be multiple roots, and they can be real or complex numbers.

Finding roots is a critical skill in mathematics as it is crucial in analyzing the properties of functions, optimizing real-world situations, and in systems of equations as illustrated in the original exercise. Identifying the roots can also lead to understanding the intercepts of a graph, which is an important aspect in the visualization of mathematical equations.