What are the solutions of the equation $(x-3)^2+2(x-3)-8=0$? Use $u$ substitution to solve.
A. $x=-5$ and $x=1$
B. $x=-1$ and $x=5$
C. $x=-1$ and $x=-7$
D. $x=1$ and $x=7$
Answer (1)
Substitute u = x − 3 into the equation, resulting in u 2 + 2 u − 8 = 0 .
Factor the quadratic equation: ( u + 4 ) ( u − 2 ) = 0 .
Solve for u : u = − 4 or u = 2 .
Substitute back to find x : x = − 1 or x = 5 .
The solutions are x = − 1 , x = 5 .
Explanation
Understanding the Problem We are given the equation ( x − 3 ) 2 + 2 ( x − 3 ) − 8 = 0 and asked to solve it using u substitution. This means we will replace the expression x − 3 with a new variable, u , to simplify the equation and make it easier to solve.
Making the Substitution Let's substitute u = x − 3 into the given equation. This gives us: u 2 + 2 u − 8 = 0
Solving for u Now we need to solve this quadratic equation for u . We can do this by factoring. We are looking for two numbers that multiply to -8 and add to 2. These numbers are 4 and -2. So we can factor the quadratic as: ( u + 4 ) ( u − 2 ) = 0
Finding the Values of u This equation is satisfied if either u + 4 = 0 or u − 2 = 0 . Solving these equations for u gives us: If u + 4 = 0 , then u = − 4 . If u − 2 = 0 , then u = 2 . So the solutions for u are u = − 4 and u = 2 .
Finding the Values of x Now we need to substitute back x − 3 for u to find the values of x . If u = − 4 , then x − 3 = − 4 . Adding 3 to both sides gives x = − 4 + 3 = − 1 . If u = 2 , then x − 3 = 2 . Adding 3 to both sides gives x = 2 + 3 = 5 . Therefore, the solutions for x are x = − 1 and x = 5 .
Final Answer The solutions to the equation ( x − 3 ) 2 + 2 ( x − 3 ) − 8 = 0 are x = − 1 and x = 5 .
Examples
Substitution is a powerful tool in mathematics that allows us to simplify complex expressions and equations. For example, imagine you are calculating the area of a garden plot that has a complex shape. By substituting simpler geometric shapes, like rectangles and triangles, you can easily calculate the area of each part and then add them up to find the total area. This approach simplifies the problem and makes it more manageable.
