Problem 45 - Intermediate Algebra

Question

Solve each quadratic equation using the method that seems most appropriate. $$ (x+2)(x-7)=10 $$

Step-by-Step Solution

Answer

The solutions are \(x = 8\) and \(x = -3\).

Step 1: Expand the Equation

First, let's expand the equation \((x+2)(x-7)=10\). Multiply the two binomials:\[(x+2)(x-7) = x^2 - 7x + 2x - 14 = x^2 - 5x - 14\]So, the equation becomes:\[x^2 - 5x - 14 = 10\]

Step 2: Rearrange to Make a Standard Quadratic Equation

Subtract 10 from both sides to set the equation to zero:\[x^2 - 5x - 14 - 10 = 0\]Which simplifies to:\[x^2 - 5x - 24 = 0\]

Step 3: Factor the Quadratic

Now, we look for two numbers that multiply to -24 and add up to -5. These numbers are -8 and 3. Hence, we can factor the quadratic as:\[(x-8)(x+3) = 0\]

Step 4: Solve for x

Using the zero-product property, set each factor equal to zero:\[\begin{align*}x-8 &= 0 \x+3 &= 0\end{align*}\]Solving these gives:\[\begin{align*}x &= 8 \x &= -3\end{align*}\]

Step 5: Verify the Solutions

Substitute both values back into the original equation to verify:For \(x = 8\):\[(8+2)(8-7) = 10(1) = 10 \text{ (True)}\]For \(x = -3\):\[(-3+2)(-3-7) = (-1)(-10) = 10 \text{ (True)}\]Hence, both solutions are correct.

Key Concepts

Factoring Quadratics

Quadratic equations typically take the form of \(ax^2 + bx + c = 0\). Factoring quadratics is a pivotal method in dealing with such equations. When factoring, we aim to write the quadratic in the form of two binomials, like \((x - p)(x - q) = 0\),where \(p\) and \(q\) are numbers that satisfy certain conditions.
Here's how you can factor a quadratic equation:

• First, identify coefficients \(a\), \(b\), and \(c\) from the standard form.
• Find two numbers that multiply to \(ac\) and add up to \(b\). This is crucial for the factoring process.
• Rewrite the middle term of the quadratic using the two numbers found.
• Finally, factor by grouping, a method to rearrange terms in bins, and simplify to get the factored form.

In our exercise, we factored \(x^2 - 5x - 24\) to \((x-8)(x+3)\), where -8 and 3 multiply to -24 and add up to -5. This is a fundamental skill in solving quadratics efficiently. Checking your work every step of the way ensures you find the correct binomial pairs.

Zero-Product Property

Once a quadratic is factored, the zero-product property comes into play. This property states that if two quantities multiply to zero, at least one of the quantities must be zero.
For the equation \((x-8)(x+3) = 0\),we apply the zero-product property by setting each factor equal to zero:

• \(x - 8 = 0\)
• \(x + 3 = 0\)

Solving these simple linear equations gives the solution set:

• \(x = 8\)
• \(x = -3\)

The zero-product property is a simple yet powerful tool because it allows us to break down more complex equations into basic ones. By isolating each factor, the solving process becomes straightforward and manageable.

Algebraic Verification

After finding the potential solutions, algebraic verification helps confirm their correctness. This process involves substituting the solutions back into the original equation to ensure they satisfy it.
To verify, substitute each solution independently:

• For \(x = 8\): Evaluate \((8+2)(8-7)\), which simplifies to \(10\times1 = 10\). This confirms the solution is correct because it equals the original right-side value.
• For \(x = -3\): Evaluate \((-3+2)(-3-7)\), yielding \((-1)\times(-10) = 10\), matching the required outcome.

Verification is an essential step as it confirms the interpretations and calculations to avoid errors. Ensuring solutions fit the original equation solidifies the understanding of solving quadratics correctly.