Solve and Check: $(a+4)^2+3(a+4)+2=0$
If there are multiple solutions, separate them using commas
Answer (1)
Substitute x = a + 4 to get x 2 + 3 x + 2 = 0 .
Factor the quadratic equation to get ( x + 1 ) ( x + 2 ) = 0 .
Solve for x to find x = − 1 or x = − 2 .
Substitute back to find a = − 5 or a = − 6 , so the answer is − 6 , − 5 .
Explanation
Simplifying the Equation We are given the equation ( a + 4 ) 2 + 3 ( a + 4 ) + 2 = 0 and asked to solve for a . This looks like a quadratic equation in disguise! Let's make a substitution to simplify things.
Substitution Let x = a + 4 . Substituting this into the original equation, we get x 2 + 3 x + 2 = 0 . Now we have a standard quadratic equation.
Factoring the Quadratic We can factor the quadratic equation as ( x + 1 ) ( x + 2 ) = 0 . This means that either x + 1 = 0 or x + 2 = 0 .
Solving for x Solving for x , we have two possible values: x = − 1 or x = − 2 .
Substituting Back Now we need to substitute back a + 4 for x to solve for a . If x = − 1 , then a + 4 = − 1 , so a = − 1 − 4 = − 5 . If x = − 2 , then a + 4 = − 2 , so a = − 2 − 4 = − 6 .
Checking the Solutions Therefore, the solutions for a are a = − 5 and a = − 6 . Let's check these solutions in the original equation. If a = − 5 , then ( − 5 + 4 ) 2 + 3 ( − 5 + 4 ) + 2 = ( − 1 ) 2 + 3 ( − 1 ) + 2 = 1 − 3 + 2 = 0 , which is correct. If a = − 6 , then ( − 6 + 4 ) 2 + 3 ( − 6 + 4 ) + 2 = ( − 2 ) 2 + 3 ( − 2 ) + 2 = 4 − 6 + 2 = 0 , which is also correct.
Final Answer The solutions are a = − 5 and a = − 6 .
Examples
Imagine you're designing a bridge and need to calculate the stress on a support beam. The equation you solved is similar to equations used to model stress and strain in materials. By finding the values of 'a' that satisfy the equation, you're essentially determining the points where the stress is at a critical level. This helps engineers ensure the bridge is safe and stable under various loads. Understanding how to solve quadratic equations is crucial for ensuring structural integrity in engineering projects.
