Find the value of each expression when [tex]x=2[/tex]:
[tex]x^2+5=[/tex] $\square$
[tex](x+5)^2=[/tex] $\square$
[tex]x^2+2x+5=[/tex] $\square$
Answer (1)
Substitute x = 2 into the expression x 2 + 5 and calculate the result: 2 2 + 5 = 4 + 5 = 9 .
Substitute x = 2 into the expression ( x + 5 ) 2 and calculate the result: ( 2 + 5 ) 2 = 7 2 = 49 .
Substitute x = 2 into the expression x 2 + 2 x + 5 and calculate the result: 2 2 + 2 ( 2 ) + 5 = 4 + 4 + 5 = 13 .
The values of the expressions at x = 2 are 9 , 49 , and 13 .
Explanation
Introduction We are given three expressions to evaluate at x = 2 . Let's take each expression one by one.
Evaluating the first expression First, we have the expression x 2 + 5 . Substituting x = 2 into this expression, we get: x 2 + 5 = ( 2 ) 2 + 5 = 4 + 5 = 9. So, the value of the first expression is 9.
Evaluating the second expression Next, we have the expression ( x + 5 ) 2 . Substituting x = 2 into this expression, we get: ( x + 5 ) 2 = ( 2 + 5 ) 2 = ( 7 ) 2 = 49. So, the value of the second expression is 49.
Evaluating the third expression Finally, we have the expression x 2 + 2 x + 5 . Substituting x = 2 into this expression, we get: x 2 + 2 x + 5 = ( 2 ) 2 + 2 ( 2 ) + 5 = 4 + 4 + 5 = 13. So, the value of the third expression is 13.
Conclusion Therefore, the values of the expressions at x = 2 are 9, 49, and 13, respectively.
Examples
Understanding how to evaluate expressions is a fundamental concept in algebra and is used in many real-world applications. For example, if you are calculating the area of a square with side length x , the area is given by x 2 . If you know that x = 5 meters, then the area is 5 2 = 25 square meters. Similarly, if you are calculating the distance traveled by an object moving at a constant speed v for a time t , the distance is given by v t . If v = 10 m/s and t = 3 seconds, then the distance is 10 × 3 = 30 meters. These simple examples show how evaluating expressions is used in everyday calculations.
