What are the solutions of the quadratic equation $(x-8)^2-13(x-8)+30=0$? Use $u$ substitution to solve.
A. $x=-11$ and $x=-18$
B. $x=-2$ and $x=5$
C. $x=2$ and $x=-5$
D. $x=11$ and $x=18
Answer (2)
Substitute u = x − 8 into the equation, resulting in u 2 − 13 u + 30 = 0 .
Factor the quadratic equation in u to get ( u − 3 ) ( u − 10 ) = 0 .
Solve for u , obtaining u = 3 or u = 10 .
Substitute back x − 8 for u and solve for x , yielding the solutions x = 11 and x = 18 . The solutions are x = 11 and x = 18 .
Explanation
Understanding the Problem We are given the quadratic equation ( x − 8 ) 2 − 13 ( x − 8 ) + 30 = 0 and asked to solve it using u -substitution. This means we will replace the expression ( x − 8 ) with a new variable, u , to simplify the equation and make it easier to solve.
Making the Substitution Let u = x − 8 . Substituting this into the original equation, we get: u 2 − 13 u + 30 = 0
Factoring the Quadratic Now we need to factor the quadratic equation u 2 − 13 u + 30 = 0 . We are looking for two numbers that multiply to 30 and add up to -13. These numbers are -3 and -10. Therefore, we can factor the equation as follows: ( u − 3 ) ( u − 10 ) = 0
Solving for u To solve for u , we set each factor equal to zero: u − 3 = 0 or u − 10 = 0 Solving these equations gives us: u = 3 or u = 10
Substituting Back and Solving for x Now we substitute x − 8 back in for u to solve for x : x − 8 = 3 or x − 8 = 10 Solving these equations gives us: x = 3 + 8 or x = 10 + 8 x = 11 or x = 18
Final Answer Therefore, the solutions to the quadratic equation are x = 11 and x = 18 .
Examples
Quadratic equations are used in various real-life situations, such as calculating the trajectory of a projectile, determining the dimensions of a rectangular area given its area and perimeter, or modeling the growth of a population. For example, if you are designing a bridge, you might use a quadratic equation to model the curve of an arch. Similarly, in business, quadratic equations can help optimize profits or minimize costs by finding the optimal production level or pricing strategy. Understanding how to solve quadratic equations is a fundamental skill in many fields.
By substituting u = x − 8 into the equation and factoring, we found that the solutions to the quadratic equation are x = 11 and x = 18 . The correct answer is option D. These solutions were derived by setting the factors of the quadratic to zero and back substituting for x .
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