Problem 45
Question
Solve each quadratic equation using the method that seems most appropriate. $$(x+2)(x-7)=10$$
Step-by-Step Solution
Verified Answer
The solutions to the equation are \(x = 8\) and \(x = -3\).
1Step 1: Expand the Equation
First, expand the left side of the equation to remove the parentheses. Use the distributive property to multiply \((x+2)(x-7)\): \[(x+2)(x-7) = x^2 - 7x + 2x - 14\]Simplify this to:\[x^2 - 5x - 14\]
2Step 2: Move All Terms to One Side
Subtract 10 from both sides of the equation to set the equation to zero:\[x^2 - 5x - 14 = 10 \rightarrow x^2 - 5x - 14 - 10 = 0 \]\[x^2 - 5x - 24 = 0\]
3Step 3: Factor the Quadratic Equation
Look for two numbers that multiply to \(-24\) and add to \(-5\). The numbers \(-8\) and \(3\) work:\[x^2 - 5x - 24 = (x - 8)(x + 3) = 0\]
4Step 4: Solve for the Roots
Set each factor equal to zero and solve for \(x\):1. \(x - 8 = 0\) \(x = 8\)2. \(x + 3 = 0\) \(x = -3\)
5Step 5: Verify the Solutions
Plug the solutions back into the original equation to verify: - For \(x = 8\): \((8+2)(8-7) = 10 \rightarrow 10(1) = 10\).- For \(x = -3\): \((-3+2)(-3-7) = 10 \rightarrow (-1)(-10) = 10\).Both solutions satisfy the original equation, confirming they are correct.
Key Concepts
Distributive PropertyFactoring QuadraticsRoots of Equations
Distributive Property
The distributive property is a fundamental algebraic principle used to eliminate parentheses by distributing a multiplier to terms within the parentheses. It is particularly helpful when expanding expressions such as \((x+2)(x-7)\). The property states that \(a(b + c) = ab + ac\).
Applying to Quadratics:
Consider \( (x + 2)(x - 7) \). By applying the distributive property:
Such simplification is essential to form a standard quadratic equation, which is key for further analysis and solving.
Applying to Quadratics:
Consider \( (x + 2)(x - 7) \). By applying the distributive property:
- First, multiply \(x \) by every term in the second parenthesis: \(x(x - 7) = x^2 - 7x\).
- Next, distribute \(2 \) the same way: \(2(x - 7) = 2x - 14\).
- Combine these results: \(x^2 - 7x + 2x - 14\).
Such simplification is essential to form a standard quadratic equation, which is key for further analysis and solving.
Factoring Quadratics
Factoring is a strategy for solving quadratic equations by expressing them as a product of simpler binomials. For example, the equation \(x^2 - 5x - 24 = 0\) can be factored.
Steps to Factor:
Steps to Factor:
- Identify the Constants: We need two numbers that multiply to \(-24\) (the constant term) and add to \(-5\) (the linear coefficient).
- Finding the Numbers: Consider the numbers \(-8\) and \(+3\). They satisfy both required conditions: \(-8 \times 3 = -24\) and \(-8 + 3 = -5\).
- Formulate the Factors: Use these numbers to factor the quadratic: \(x^2 - 5x - 24 = (x - 8)(x + 3)\).
Roots of Equations
The solutions to a quadratic equation, often referred to as the roots, are the values of \(x\) that satisfy the equation \(ax^2+bx+c=0\).
Finding Roots:
Verification:
Verifying the roots by substituting back into the original equation is crucial to ensure they are correct. For instance:
Finding Roots:
- Once the quadratic is factored, use each factor to set up an equation: \(x - 8 = 0\) and \(x + 3 = 0\).
- Solve for \(x\):
- For \(x - 8 = 0\), you find \(x = 8\).
- For \(x + 3 = 0\), you find \(x = -3\).
Verification:
Verifying the roots by substituting back into the original equation is crucial to ensure they are correct. For instance:
- For \(x = 8\), substitute to get \((8 + 2)(8 - 7) = 10 \).
- For \(x = -3\), substitute to check \((-3 + 2)(-3 - 7) = 10 \).
Other exercises in this chapter
Problem 45
Set up an equation and solve each problem. Find two numbers such that their sum is 10 and their product is 22 .
View solution Problem 45
Use the quadratic formula to solve each of the quadratic equations. Check your solutions by using the sum and product relationships. $$12 x^{2}-73 x+110=0$$
View solution Problem 45
Write each of the following in terms of \(i\), perform the indicated operations, and simplify. For example, $$ \begin{aligned} \sqrt{-3} \sqrt{-8} &=(i \sqrt{3}
View solution Problem 46
Solve each inequality. $$2 x^{3}+4 x^{2} \leq 0$$
View solution