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Question 1176836: . Find four rational numbers such that the product of the first, third, and fourth numbers is --6. The second number is 3 less than the first number, the third is 2 less than the second, and the fourth is 2 less than the third
Found 3 solutions by math_tutor2020, MathLover1, greenestamps:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
f = first number
s = second number
t = third number
h = fourth number
the letter f was already taken, so I picked the last letter in "fourth"
All of these values are some rational number in the form a/b, where a,b are integers and b is nonzero.
We are given these four facts- The product of the first, third, and fourth numbers is -6.
- The second number is 3 less than the first number
- The third is 2 less than the second
- The fourth is 2 less than the third
Those facts lead to the corresponding equations (in the order shown)
f*t*h = -6
s = f-3
t = s-2
h = t-2
I'll refer to them as equations (1) through (4)
Let's start with equation (4)
h = t-2
Now replace t with s-2; this is valid due to equation (3)
h = t-2
h = (t)-2
h = (s-2)-2
h = s-4
This tells us the fourth value (h) is 4 less than the second value (variable s).
We can add 4 to both sides getting
s = h+4
Which then can be plugged into equation (2)
s = f-3
h+4 = f-3
h = f-3-4
h = f-7
The fourth value is 7 less than the first value
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We have these equations all with h to start off with
h = f-7
h = s-4
h = t-2
I sorted them so that the one with 'f' goes first, then 's' second, and 't' third
For each of those three equations shown, isolate the other variable. Doing so leads to these three results
f = h+7
s = h+4
t = h+2
What this allows us to do is replace the f and t values in equation (1) with something in terms of h. That way we can set up a single variable equation in which we can solve for h
f*t*h = -6
(f)*(t)*h = -6
(h+7)*(h+2)*h = -6
(h^2+9h+14)*h = -6 ..... FOIL rule
h^3+9h^2+14h = -6 .... distribute
h^3+9h^2+14h+6 = 0
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Recall that each f,s,t,h are all rational numbers.
That must mean the last equation we arrived at has at least one rational root; or else, all the roots of that cubic equation are irrational and that contradicts the instructions.
The rational root theorem says that we divide the factors of the last term (6) over the factors of the leading coefficient (1)
List out the plus and minus versions of each ratio
Luckily that list isn't too big
If the numerator is 6, then we have the ratios: -6/1, 6/1
That leads to -6 and 6 respectively.
If the numerator is 3, then we get -3/1 = -3 and 3/1 = 3
If the numerator is 2, then we get -2/1 = -2 and 2/1 = 2
If the numerator is 1, then we get -1/1 = -1 and 1/1 = 1
The possible rational roots are:
-6,6
-3,3
-2,2
-1,1
As you can see, all we've done really is listed all the factors of 6 (positive and negative versions). This is due to the leading coefficient being 1.
From here, we plug each of those 8 values into the cubic equation
h^3+9h^2+14h+6 = 0
to see if the left side becomes 0 or not
Let's try out h = -6
h^3+9h^2+14h+6 = (-6)^3+9*(-6)^2+14*(-6)+6 = 30
That doesn't work.
Now try h = 6
h^3+9h^2+14h+6 = (6)^3+9*(6)^2+14*(6)+6 = 630
That doesn't work either.
Through trial and error, you should find that only h = -1 works since,
h^3+9h^2+14h+6 = (-1)^3+9*(-1)^2+14*(-1)+6 = 0
So h = -1 is a rational root solution to h^3+9h^2+14h+6 = 0
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With that in mind, we can find the other values
f = h+7 = -1+7 = 6
s = h+4 = -1+4 = 3
t = h+2 = -1+2 = 1
To summarize, we have these values:
f = 6
s = 3
t = 1
h = -1
as the first through fourth values in that order.
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Check:
We'll plug in the values we found for each equation formed at the top of this page.
f*t*h = -6
6*1*(-1) = -6
-6 = -6
First equation works out
s = f-3
3 = 6-3
3 = 3
That works out also
t = s-2
1 = 3-2
1 = 1
Works too
h = t-2
-1 = 1-2
-1 = -1
We can see that all four original equations work; therefore, the answers have been confirmed.
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Answers:
first = 6
second = 3
third = 1
fourth = -1
Answer by MathLover1(20849)  (Show Source):
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
You can ignore one of the responses you have received, since it answers a different question than the one that is asked.
I would advise you also to ignore most of the response from the other tutor, who takes 20-30 steps to start with four unknowns and four equations to end up with a single equation in one unknown. That might be good exercise in algebra, but it is a huge waste of time.
Take the time (you don't need much) to set up the problem using a single variable:
1st number: x
2nd number: x-3 (3 less than the 1st)
3rd number: x-5 (2 less than the 2nd)
4th number: x-7 (2 less than the 3rd)
Given that the product of the 1st, 3rd, and 4th is -6, we have
Solving this algebraically means finding the roots of a cubic equation, which is a lengthy process. Or you could finish solving the problem using a graphing calculator.
But, assuming a solution is to be obtained without a calculator, I would solve the problem from here using logical reasoning and some simple calculations.
The equation says the product of x, x-5, and x-7 is a negative number. Simple analysis shows that the product is negative for values of x less than 0, or between 5 and 7. Negative answers are far less likely, so we can look for an answer between 5 and 7.
Then, since the product is EXACTLY -6, it is very likely that the factors x, x-5, and x-7 are all whole numbers. So we try x=6 and find x(x-5)(x-7) = 6(1)(-1) = -6, as we wanted.
ANSWERS:
1st number = x = 6
2nd number = x-3 = 3
3rd number = x-5 = 1
4th number = x-7 = -1
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