Problem 118 - College Algebra Essentials

Question

In Exercises \(115-122,\) find all values of \(x\) satisfying the given conditions. $$ y_{1}=x-3, y_{2}=x+8, \text { and } y_{1} y_{2}=-30 $$

Step-by-Step Solution

Answer

The solution to the problem is \(x = -2\) and \(x = -3\).

Step 1: Substitute \(y_{1}\) and \(y_{2}\) with their expressions

Instead of \(y_{1} y_{2}\) in the third equation, use \(y_{1} = x-3\) and \(y_{2} = x+8\). This turns the equation \(y_{1} y_{2}=-30\) into \((x-3)(x+8) = -30\).

Step 2: Simplify the Equation

Expand the product on the left hand side of the equation: \(x^2 + 5x - 24 = -30\).

Step 3: Solve for \(x\)

Now solve the quadratic equation for \(x\). First make the right side 0 by adding 30 to both sides: \(x^2 + 5x + 6 = 0\). Now find the roots of the equation by factoring the left hand side into \((x+2)(x+3)=0\). Setting each factor equal to zero gives \(x = -2\) and \(x = -3\).

Key Concepts

Factoring

Factoring is a method used to simplify algebraic expressions by breaking them down into simpler components called "factors." In the context of a quadratic equation, factoring can help us solve for the unknown variable.

• A quadratic equation typically takes the form: \( ax^2 + bx + c = 0 \).
• To factor it, we look for two numbers that multiply to \( a \times c \) and add to \( b \).

By expressing a quadratic equation as a product of linear factors, it becomes easier to find the solutions. For instance, in our example, we need to solve \( x^2 + 5x + 6 = 0 \) by finding two numbers that multiply to 6 and add up to 5. The numbers 2 and 3 fit this requirement. Therefore, we can factor the equation as \((x+2)(x+3)=0\).Factoring is particularly useful in solving quadratic equations because once in this form, we can proceed to identify the roots directly by setting each factor to zero.

Roots of Equations

The roots of an equation are the values of the variable that satisfy the equation, making it equal to zero. Finding the roots is a crucial part of solving polynomial equations, like quadratics.

• For the quadratic equation \( ax^2 + bx + c = 0 \), the roots can often be found by factoring, using the quadratic formula, or completing the square.
• Once factored, as in our exercise example of \((x+2)(x+3)=0\), each factor set to zero gives the possible roots.

Hence, setting \( x+2 = 0 \) results in \( x = -2 \), and setting \( x+3 = 0 \) results in \( x = -3 \). These values are the roots, meaning if plugged back into the original equation, they satisfy it completely making the equation equal zero.

Polynomial Equations

Polynomial equations are algebraic expressions that involve powers of the variable. Quadratics, like \(x^2 + 5x + 6 = 0\), are polynomial equations often with a degree of two.

• These equations can often be solved using techniques such as factoring, the quadratic formula, or graphing.
• Understanding polynomial equations is essential because they frequently occur in both pure and applied mathematics.

The equation in our example, \( (x-3)(x+8) = -30 \), is a rearranged version of a quadratic polynomial. By expanding and simplifying it, we achieve \(x^2 + 5x + 6 = 0\), allowing us to find meaningful solutions, often called roots, as demonstrated with factoring. As students become comfortable with these manipulations, they gain a critical tool for solving more complex mathematical problems.