SOLUTION: the square root of th sum of two consecutive odd integers is 6√2. Find the larger integer.

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Question 945304: the square root of th sum of two consecutive odd integers is 6√2. Find the larger integer.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
2n+1 and 2n+3 are the two integers.

sqrt%282n%2B1%2B2n%2B3%29=6sqrt%282%29

sqrt%284n%2B4%29=6sqrt%282%29

2sqrt%28n%2B1%29=2%2A3%2Asqrt%282%29

sqrt%28n%2B1%29=3%2Asqrt%282%29

What do you guess comes next?




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The integers are 35 and 37.

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

the square root of th sum of two consecutive odd integers is 6√2. Find the larger integer.
Let larger integer be L

Then smaller is: L - 2

We then have: sqrt%28L+%2B+L+-+2%29+=+6sqrt%282%29

sqrt%282L+-+2%29+=+6sqrt%282%29

%28sqrt%282L+-+2%29%29%5E2+=+%286sqrt%282%29%29%5E2 ------- Squaring both sides

2L - 2 = 36(2)

2L - 2 = 72

2L = 72 + 2

2L = 74

L, or larger integer = 74%2F2, or highlight_green%2837%29      

You can do the check!! 

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