Problem 35 - College Algebra Essentials

Question

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

Step-by-Step Solution

Answer

The roots of the equations are \(x = 0, y = 0\) and \(x = -1, y = 1\)

Step 1: Simplify the Second Equation

Isolate \(y\) in the second equation, which can be rewritten as: \(y = x^{2}\)

Step 2: Substitute the Result into the First Equation

Substitute \(y = x^{2}\) into the first equation. The first equation becomes: \(x^{3} + x^{2} = 0\)

Step 3: Solve for \(x\)

Factorize the equation to solve for \(x\). Factorizing the equation results in: \(x^{2} (x + 1) = 0\). Solving this equation gives roots \(x = 0\) and \(x = -1\)

Step 4: Find the Corresponding \(y\)-values

Substitute the found \(x\)-values into \(y = x^{2}\) to find the corresponding \(y\)-values. Substituting \(x = 0\) results in \(y = 0\), and substituting \(x = -1\) results in \(y = 1\)

Key Concepts

Substitution Method

The substitution method is a technique used in solving systems of equations where one solves one equation for one variable and substitutes that expression into another equation. This method is beneficial as it reduces the number of variables in the given system. Here is how the substitution method works step-by-step:

• Solve for one variable: Start by solving one of the equations for one of the variables. For example, in the system \(\{\begin{array}{l} x^{3}+y=0 \ x^{2}-y=0 \end{array}\},\) we solve the second equation for \(y\): \(y = x^{2}\).
• Substitute into the other equation: Substitute this expression for \(y\) into the first equation. Hence, \(x^{3} + x^{2} = 0\).
• Solve the resulting equation: After substitution, you do not have a system anymore, just one equation in one variable, making it easier to solve.

This method is particularly helpful because it can simplify the process significantly when equations are otherwise challenging to manipulate directly.

Cubic Equations

Cubic equations, like \(x^{3} + x^{2} = 0\), are polynomials of degree three. Solving a cubic equation involves finding the values of \(x\) that satisfy the equation. Here are some key steps and techniques:

• Identify the degree: Recognize that an equation's degree is the highest exponent on the variable, which in this case is 3.
• Method of factoring: If possible, factor out common terms. For \(x^{3} + x^{2} = 0\), you can factor out \(x^{2}\), resulting in \(x^{2}(x + 1) = 0\).
• Roots of the equation: After factorization, solve each component separately. Here, either \(x^{2} = 0\) or \(x + 1 = 0\), providing solutions \(x = 0\) and \(x = -1\).

These steps are crucial for breaking down and understanding cubic equations, especially when they appear in a system of equations.

Quadratic Equations

Quadratic equations are polynomials of degree two, commonly structured as \(ax^{2} + bx + c = 0\). Solving them often involves using specific strategies to find their roots. In the given exercise, the equation for \(y\) transforms into a quadratic form:

• Standard form: The equation \(y = x^{2}\) is essentially quadratic, though simplified in this context to illustrate a basic relation.
• Finding roots: In scenarios where quadratic equations become more complex, like \(x^{2} + bx + c = 0\), methods such as factoring, completing the square, or the quadratic formula \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) would be employed.
• Using roots in systems: In systems of equations, solving quadratics helps determine one variable's value based on another known variable, simplifying the resolution of the entire system.

Quadratic equations appear frequently in mathematical problems and understanding their properties simplifies solving systems of equations.

Factoring Polynomials

Factoring polynomials is a method of expressing a polynomial as a product of its factors, which are simpler expressions that, when multiplied together, give the original polynomial. Let's break this down further:

• Identify common factors: Begin by finding the greatest common factor (GCF) in the polynomial terms. For example, in \(x^{3} + x^{2} = 0\), the GCF is \(x^{2}\).
• Re-write the equation: Factor out the common term. After factoring \(x^{2}\) from \(x^{3} + x^{2}\), the expression becomes \(x^{2}(x + 1) = 0\).
• Solve for roots: Once factored, set each factor equal to zero and solve to find the roots. Here, we solve \(x^{2} = 0\) and \(x + 1 = 0\), resulting in \(x = 0\) and \(x = -1\).

Factoring efficiently simplifies polynomials, making seemingly complex equations manageable and allowing for quick, accurate solutions.