SOLUTION: 1.Find three consecutives odd integers such that the product of the first and third integers is 4 less than the square of the second integer. 2.The product of the two-digit no.

Question 164204: 1.Find three consecutives odd integers such that the product of the first and third integers is 4 less than the square of the second integer.
2.The product of the two-digit no. and its tens digit is 54. Find the no. if the sum of the digits when added to the no. gives a result of 36.
i like math but it's hard(lol)!!
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
1. Find three consecutives odd integers such that the product of the first and
third integers is 4 less than the square of the second integer.
:
Three odd integers: x, (x+2), (x+4)
:
The product of 1st & 3rd, is 4 less than the 2nd squared
x(x+4) = (x+2)^2 - 4
:
x^2 + 4x = x^2 + 4x + 4 - 4
x^2 + 4x = x^2 + 4x
Which is, of course true, but means there is not a unique solution
Pick any 3 odd consecutive numbers and this equation will be true
(It's also true for 3 consecutive even numbers)
Try it on your calc.
:
:
2.The product of the two-digit no. and its tens digit is 54. Find the no.,
if the sum of the digits when added to the no. gives a result of 36.
:
Let = 10x + y = the two digit number
:
10's digit times the number = 54
x(10x+y) = 54
10x^2 + xy = 54
:
number + sum of the digits = 36
(10x+y) + x + y = 36
11x + 2y = 36
2y = 36 - 11x
y = (18 - 5.5x); divided the equation by 2
:
10x^2 + x(18-5.5x) = 54
10x^2 + 18x - 5.5x^2 = 54
10x^2 - 5.5x^2 + 18x = 54
4.5x^2 + 18x - 54 = 0; multiplied equation to get rid of the decimal
9x^2 + 36x - 108 = 0
simplify, divide by 9
x^2 + 4x - 12 = 0
(x + 6)(x - 2) = 0
x = -6
x = +2
:
Let's just use the positive solution here; x = 2:
:
Find y using the equation: x(10x+y) = 54, substitute 2 for x
2(10(2) + y)= 54
2(20 + y) = 54
40 + 2y = 54
2y = 54 - 40
2y = 14
y = 7
:
The "number" is 27
:
you can check the solutions in the two original equations

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