Question 586686: Question 1: Twice the product of 2 consecutive integers.
Question 2: Seven less than twice a number is 9 more than the number.
Question 3: The product of 3 consecutive even integers is eight times the largest number.
Answer by Sudoku19(4)   (Show Source):
You can put this solution on YOUR website! Question 1 : Twice the product of 2 consecutive integers.
Let say x is the integer we have. The next number to it is called the consecutive number. So, the next number would be x+1.
The product of 2 consecutive integers is x(x + 1). The twice of it is multiplying the equation into two. So, the equation would be 2[x(x + 1)]. When simplified, the answer would be: 2x^2 + 2x
--------------------------------------------------------------------------------
Question 2: Seven less than twice a number is 9 more than the number.
Again x will be our number. Seven less than twice the number is portrayed as 7 - 2x. 9 more than the number is given as 9 + x. The equation would be: 7 - 2x = 9 + x. The answer is: x = -2/3 or 2/-3
--------------------------------------------------------------------------------
Question 3: The product of 3 consecutive even integers is eight times the largest number.
Same variable for the expression. Now we have consecutive even integers. Would it be same as question number one? NO! Consecutive even integers! This is counted by twos. It is counted as x, x + 2, x + 4. So the product of 3 consecutive even integers is written as x[x + 2(x + 4)] and the expression, eight times the largest number, would be 8(x + 4). Why "x + 4" you say? Since the highest possible number would be the x + 4, because it will receive the highest addition of 4. Anyway, the equation would be x[x + 2(x + 4)] = 8(x + 4). The answer would be x^3 + 6x^2 - 32 = 0.
|
|
|
