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The sum of the squares of 2 consecutive negative integers is 41. What are the numbers?
Which of the following equations is the result of using the factoring method to solve the problem?

 

(n - 5)(n - 4) = 0

(n - 5)(n + 4) = 0

(n + 5)(n - 4) = 0

(n + 5)(n + 4) = 0

 

There are two factors of -36 such that one factor is 11 less than half of the other factor. Choose all the pairs of these factors.

 

-2 and 18

-6 and 6

3 and -12

4 and -9

 Jan 18, 2019
 #1
avatar+111357 
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First one....we have that

 

n^2 + ( n + 1)^2  =  41

 

n^2 + n^2 + 2n + 1 = 41

 

2n^2 + 2n - 40  = 0

 

n^2 + n  - 20 =  0

 

(n  + 5) ( n - 4) = 0

 

Second one

 

Call the first factor, F

So...the second is   F/2 - 11

 

So.....this implies that

 

(F)  [ (F/2) - 11]   = -36       simplify

 

F^2/2 - 11F = - 36

 

F^2/2 - 11F + 36 = 0         multiply through by 2

 

F^2 - 22F + 72 = 0

 

(F - 18) ( F - 4)  = 0

 

F = 18      or F = 4

 

When F = 18    the other factor is :  F/2 - 11   =  18/2 - 11  =  9 - 11 =   -2

When F = 4    the other factor is   4/2 - 11 =  2 - 11  =  -9

 

So....the factors are

 

(18, -2)   and ( 4, - 9)

 

 

cool cool cool

 Jan 18, 2019
 #2
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0

Which of the following equations is the result of using the factoring method to solve the problem?

 

(n - 5)(n - 4) = 0

(n - 5)(n + 4) = 0

(n + 5)(n - 4) = 0

(n + 5)(n + 4) = 0

 

 

(n - 5)(n - 4)  = 0,     This comes from this quadratic:  n^2 - 9 n + 20 = 0
(n - 5)(n + 4) = 0,     This comes from this quadratic:  n^2 - n - 20 = 0
(n + 5)(n - 4) = 0,     This comes from this quadratic:  n^2 + n - 20 = 0
(n + 5)(n + 4) =0,     This comes from this quadratic:  n^2 + 9 n + 20 = 0

 Jan 18, 2019

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