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Question 165031: 1. Find two positive integers such that one is three times the other and their product is 18 more than the larger
2. 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.
3. 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.
4. Find a three digit no. whose ten’s digit is 4 times the hundred’s digit and the unit’s digit is one more than the ten’s digit. The square of the sum of the digits is 72 more than the no?
can you show how you equate this? please and thank you!!!
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 1. Find two positive integers such that one is three times the other and their product is 18 more than the larger
:
let the two integers be x, 3x
an equation for what it says:
x*3x = 3x + 18
3x^2 - 3x - 18 = 0; arrange as a quadratic equation
Simplify, divide equation by 3
x^2 - x - 6 = 0
Factors to
(x-3)(x+2) = 0
Two solutions
x = +3, the larger number is 9
and
x = -2, the larger number is -6
":
Check the 1st solution: 3*9 = 9 + 18
Check the 2nd solution: -2*-6 = -6 + 18
;
:
2. Find three consecutive 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)
x*(x+4) = (x+2)^2 - 4
x^2 + 4x = x^2 + 4x + 4 - 4
x^2 + 4x = x^2 + 4x
You can see there is no unique solution, x can be any value
:
:
3. 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 the two digit number = 10x + y
:
"product of the two-digit no. and its tens digit is 54." equation for this is:
(10x + y) * x = 54
:
"Find the no. if the sum of the digits when added to the no. gives a result of "36.
x + y + (10x + y = 36
4. Find a three digit no. whose
:
The 3 digit number: (100x + 10y + z)
:
"ten’s digit is 4 times the hundred’s digit"
y = 4x
or
x = .25y
:
"the unit’s digit is one more than the ten’s digit."
z = (y+1)
:
"The square of the sum of the digits is 72 more than the no?"
(x+y+z)^2 = 100x + 10y + z + 72
:
Substitute .25y for x, and (y+1) for z in the above equation:
(.25y+y+(y+1)z)^2 = 100(.25y) + 10y + (y+1) + 72
:
FOIL on the left Combine like terms on the right:
(2.25y+1)^2 = 25y + 10y + y + 1 +72
:
5.0625y^2 + 4.5y + 1 = 36y + 73
:
Combine like terms on the left
5.0625y^2 + 4.5y - 36y + 1 - 73 = 0
:
5.0625y^2 - 31.5y - 72 = 0
:
A quadratic equation, we have to solve with the quadratic formula:
a = 5.0625; b = -31.5; c = -72
Assuming you know how to do this:
the positive solution: y = 8
:
It's easy to find the other two digits, I'll let you do that.
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