SOLUTION: Please help. I need to find three consectutive integers such that the sum of their squares is 77. Thanks!!

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Question 87338: Please help. I need to find three consectutive integers such that the sum of their squares is 77. Thanks!!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let x=1st #, y=2nd #, z=3rd #

x%5E2%2By%5E2%2Bz%5E2=77 This is the translation of "the sum of their (each number) squares is 77"

Since each number is consecutive, we're really adding 1 each time. So

y=x%2B1 and z=%28x%2B1%29%2B1=x%2B2


x%5E2%2B%28x%2B1%29%5E2%2B%28x%2B2%29%5E2=77 Plug in y=x%2B1 and z=x%2B2

x%5E2%2B%28x%5E2%2B2x%2B1%29%2B%28x%5E2%2B4x%2B4%29=77 Foil

3x%5E2%2B6x%2B5=77 Combine like terms

3x%5E2%2B6x%2B5-77=0 Subtract 77 from both sides

3x%5E2%2B6x-72=0 Combine like terms

Now lets use the quadratic formula to solve for x


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general form of the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29

So lets solve 3%2Ax%5E2%2B6%2Ax-72=0

x+=+%28-6+%2B-+sqrt%28+%286%29%5E2-4%2A3%2A-72+%29%29%2F%282%2A3%29 Plug in a=3, b=6, and c=-72



x+=+%28-6+%2B-+sqrt%28+36-4%2A3%2A-72+%29%29%2F%282%2A3%29 Square 6 to get 36



x+=+%28-6+%2B-+sqrt%28+36%2B864+%29%29%2F%282%2A3%29 Multiply -4%2A-72%2A3 to get 864



x+=+%28-6+%2B-+sqrt%28+900+%29%29%2F%282%2A3%29 Combine like terms in the radicand (everything under the square root)



x+=+%28-6+%2B-+30%29%2F%282%2A3%29 Simplify the square root



x+=+%28-6+%2B-+30%29%2F6 Multiply 2 and 3 to get 6

So now the expression breaks down into two parts

x+=+%28-6+%2B+30%29%2F6 or x+=+%28-6+-+30%29%2F6

Lets look at the first part:

x=24%2F6 Add the terms in the numerator
x=4 Divide

So one answer is
x=4
Now lets look at the second part:

x=-36%2F6 Subtract the terms in the numerator
x=-6 Divide

So another answer is
x=-6

So our solutions are:
x=4 or x=-6

Notice when we graph 3%2Ax%5E2%2B6%2Ax-72 we get:

+graph%28+500%2C+500%2C+-16%2C+14%2C+-16%2C+14%2C3%2Ax%5E2%2B6%2Ax%2B-72%29+

and we can see that the roots are x=4 and x=-6. This verifies our answer

So our first number could be 4 or -6

Lets use x=4 to find the 2nd and 3rd number
y=4%2B1 and z=4%2B2
y=5 and z=6
So we have 3 numbers: 4,5,6

Check:
4%5E2%2B5%5E2%2B6%5E2=77
16%2B25%2B36=77
77=77 works

So the 3 numbers 4,5,6 work


Now lets use x=-6 to find the 2nd and 3rd number
y=-6%2B1 and z=-6%2B2
y=-5 and z=-4
So we have 3 numbers: -6,-5,-4

Check:
%28-6%29%5E2%2B%28-5%29%5E2%2B%28-4%29%5E2=77
36%2B25%2B16=77
77=77 works

So the 3 numbers -6,-5,-4 work

Answer:
So if you allow negative numbers, you get 2 possible answers
4,5,6
or
-6,-5,-4