SOLUTION: The product of three integers is 90. The second number is twice the first number. The third number is two more than the first number. What are the three numbers? Use a polynomial a
Question 1177626: The product of three integers is 90. The second number is twice the first number. The third number is two more than the first number. What are the three numbers? Use a polynomial and the calculator to answer the question. Found 4 solutions by mananth, josgarithmetic, greenestamps, ikleyn:Answer by mananth(16946) (Show Source): You can put this solution on YOUR website! The product of three integers is 90.
let the numbers be x,y,z
The second number is twice the first number.
y = 2x
The third number is two more than the first number.
z = x+2
Substitute for y and for z and solve for x
x(2x)(x+2) = 90
2x^3+4x^2-90 = 0
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x^3 + 2x^2 - 45 = 0-
graph the cubic and you will get a solution x = 3
y = 2x = 6
z = x+2 = 5 Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
Formal algebra doesn't really help to solve this problem.
One response you have received says form the equation and look for solutions informally.
The other response says graph the equation to find the solution.
Both solution methods are valid; but the formal algebra parts of the solutions didn't get you any closer to the answer -- you still had to "look for" the answer, or get it by using a graphing utility.
This problem is best solved informally from the beginning, without trying to use formal algebra.
The prime factorization of 90 is 2*3*3*5. Combine two of those factors in a way that gives you three integers that satisfy the conditions of the problem. The way to do that is easily found: (2*3)*3*5 = 6*3*5. 6 is twice 3; and 5 is 2 more than 3.