Problem 46 - Intermediate Algebra

Question

Solve each quadratic equation using the method that seems most appropriate. $$ (x-3)(x+5)=-7 $$

Step-by-Step Solution

Answer

The solutions are \(x = -4\) and \(x = 2\).

Step 1: Expand the Quadratic Equation

First, we need to expand the given equation. Start by applying the distributive property to the left side: \((x-3)(x+5)\). This results in: \[x^2 + 5x - 3x - 15 = x^2 + 2x - 15\] Now substitute this into the original equation: \[x^2 + 2x - 15 = -7\]

Step 2: Move All Terms to One Side

Add 7 to both sides of the equation to move all terms to one side and make the right side equal to 0: \[x^2 + 2x - 15 + 7 = 0\] Simplify the equation: \[x^2 + 2x - 8 = 0\]

Step 3: Factor the Quadratic Equation

Now, factor the quadratic equation \[x^2 + 2x - 8 = 0\]. Look for two numbers that multiply to -8 and add to 2. These numbers are 4 and -2. Rewrite the equation as a product of two binomials: \[(x + 4)(x - 2) = 0\]

Step 4: Solve for x

Set each factor equal to zero to find the solutions for \(x\): First equation: \[x + 4 = 0\] Second equation: \[x - 2 = 0\] Solve each equation: \[x = -4\] and \[x = 2\]

Step 5: Verify the Solutions

We should verify our solutions by substituting them back into the original equation. For \(x = -4\): \((-4 - 3)(-4 + 5) = (-7)(1) = -7\), which is true. For \(x = 2\): \((2 - 3)(2 + 5) = (-1)(7) = -7\), which is also true.

Key Concepts

Factoring Quadratics

Factoring quadratics is like finding the "hidden products" within a quadratic equation. Quadratic equations often have the form \( ax^2 + bx + c = 0 \). To factor them, we're looking for two binomials whose product gives us the original quadratic. This method is especially useful when the quadratic equation can be quickly transformed into these binomials, making it easier to solve.

For example, given the equation \( x^2 + 2x - 8 = 0 \), we identify two numbers that multiply to \(-8\) and add up to \(2\). These numbers are \(4\) and \(-2\). Thus, we can express the quadratic as \((x + 4)(x - 2) = 0\).

By setting each binomial to zero, we find the solutions of the quadratic equation. Factoring simplifies what could be a more complicated operation into a straightforward multiplication problem.

When the numbers work out well, factoring is often the fastest and simplest method.

Distributive Property

The distributive property is a fundamental principle in algebra. It helps us simplify expressions and perform operations with ease. This property states that for any numbers \(a\), \(b\), and \(c\), \( a(b + c) = ab + ac \). In simple terms, you "distribute" the \(a\) across the \(b\) and \(c\).

This property is crucial when expanding quadratic equations. For example, expanding \((x-3)(x+5)\) involves applying the distributive property:

• Multiply \(x\) by \(x\), get \(x^2\)
• Multiply \(x\) by \(5\), get \(5x\)
• Multiply \(-3\) by \(x\), get \(-3x\)
• Multiply \(-3\) by \(5\), get \(-15\)

Combining these, you have \(x^2 + 5x - 3x - 15\), which simplifies to \(x^2 + 2x - 15\).

The distributive property breaks the multiplication into smaller, manageable parts, making it easier to solve complex problems.

Verifying Solutions

Verifying solutions of quadratic equations ensures that the answers we obtain are correct. It's like "double-checking" our work. By substituting the solutions back into the original equation, we can confirm they satisfy the equation.

For example, let's verify the solutions \(x = -4\) and \(x = 2\) for \((x-3)(x+5) = -7\).

• For \(x = -4\): Substitute into the equation to check: \((-4-3)(-4+5) = (-7)(1) = -7\). This is true.
• For \(x = 2\): Substitute again: \((2-3)(2+5) = (-1)(7) = -7\). This is also true.

Both checks hold, confirming our solutions are correct.

Verification is an essential step in solving problems accurately, affirming that our work has been done correctly and thoroughly.