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Chapter 5: Problem 11

Solve each equation, and check the solutions. $$ (x+5)(x-2)=0 $$

Short Answer

Expert verified
The solutions are \( x = -5 \) and \( x = 2 \).

Step by step solution

01

Identify the factors

Given the equation \( (x+5)(x-2)=0 \), we have the factors of the quadratic equation: \( (x+5) \) and \( (x-2) \).
02

Set each factor to zero

To find the values of \(x\), set each factor equal to zero. \[ x+5=0 \] and \[ x-2=0 \]
03

Solve for x

Solve each equation separately: \[ x+5=0 \Rightarrow x=-5 \] and \[ x-2=0 \Rightarrow x=2 \]
04

Check the solutions

Substitute \(x=-5\) and \(x=2\) back into the original equation \( (x+5)(x-2)=0 \) to verify the solutions. \ For \ x=-5 \: \ ((-5)+5)((-5)-2)=0 \Rightarrow 0 \cdot (-7) = 0 \For \ x=2 \: \ (2+5)(2-2) = 0 \Rightarrow 7 \cdot 0 = 0 \Both satisfy the equation, so the solutions are correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factoring
Factoring is one of the key methods used to solve quadratic equations. It involves breaking down a quadratic expression into simpler 'factors' whose product is the original expression. For instance, in the equation \((x+5)(x-2)=0\), the expression \((x+5)(x-2)\) is the factored form of a quadratic equation. The goal is to write the quadratic in a form where it looks like \((ax+b)(cx+d)=0\). This makes it easier to solve for the variable, because each factor can individually be set to zero.
Zero Product Property
The zero product property states that if the product of two numbers is zero, then at least one of the numbers must be zero. This property is essential when solving factored quadratic equations. Given the equation \((x+5)(x-2)=0\), the zero product property allows us to set each factor to zero: \(x+5=0\) and \(x-2=0\). By solving these simpler equations, we find the roots of the quadratic equation, which are the values of \(x\) that make the equation true.
Checking Solutions
Checking solutions is a crucial step to verify that the values obtained indeed satisfy the original equation. After solving \(x+5=0\) and \(x-2=0\), we get \(x=-5\) and \(x=2\). To check these solutions, substitute them back into the original equation \((x+5)(x-2)=0\).
For \(x=-5\):
Substitute \(-5\) into the equation: \[(-5+5)(-5-2) = 0\]
This simplifies to \(0 \cdot (-7) = 0\), which is true.
For \(x=2\):
Substitute \(2\) into the equation: \[(2+5)(2-2) = 0\]
This simplifies to \(7 \cdot 0 = 0\), which is also true.
Both solutions satisfy the original equation, confirming that they are correct.

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