document.write( "Question 81771: Problem
\n" ); document.write( "Find three consecutive integers such that the sum of their squares is 77.
\n" ); document.write( "This is what I have, but not sure.
\n" ); document.write( "x^2 + x^2 + x^2 = 77 I don't know where to go from here. Thank you for your help.
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Algebra.Com's Answer #58595 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Let x=1st #, y=2nd #, z=3rd #\r
\n" ); document.write( "\n" ); document.write( "Since you have consecutive integers, this means the integers are right next to each other. For instance, 1,2, and 3 are all consecutive integers. If x=1, then y=2, and z=3. So if you wanted them all in terms of x, you would say:\r
\n" ); document.write( "\n" ); document.write( "x=1, y=x+1, z=x+2\r
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\n" ); document.write( "\n" ); document.write( "Since if you plug in x=1 into x+2, you get z=3. So the same applies to this problem. \r
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\n" ); document.write( "\n" ); document.write( "So now we're adding the squares of these numbers to get 77 like this:\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2%2By%5E2%2Bz%5E2=77\"\r
\n" ); document.write( "\n" ); document.write( "But we can write y as \"x%2B1\" and z as \"x%2B2\" so we can replace y and z:\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2%2B%28x%2B1%29%5E2%2B%28x%2B2%29%5E2=77\"\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2%2Bx%5E2%2B2x%2B1%2Bx%5E2%2B4x%2B4=77\" Foil the two parenthesis separately\r
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\n" ); document.write( "\n" ); document.write( "\"3x%5E2%2B6x%2B5=77\" Combine like terms\r
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\n" ); document.write( "\n" ); document.write( "\"3x%5E2%2B6x%2B5-77=cross%2877-77%29\" Subtract 77 from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"3x%5E2%2B6x-72=0\"\r
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\n" ); document.write( "\n" ); document.write( "Now that we have a quadratic, lets use the quadratic formula to solve for x:\r
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation \"ax%5E2%2Bbx%2Bc=0\" (in our case \"3x%5E2%2B6x%2B-72+=+0\") has the following solutons:
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\n" ); document.write( " \"x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"
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\n" ); document.write( " For these solutions to exist, the discriminant \"b%5E2-4ac\" should not be a negative number.
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\n" ); document.write( " First, we need to compute the discriminant \"b%5E2-4ac\": \"b%5E2-4ac=%286%29%5E2-4%2A3%2A-72=900\".
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\n" ); document.write( " Discriminant d=900 is greater than zero. That means that there are two solutions: \"+x%5B12%5D+=+%28-6%2B-sqrt%28+900+%29%29%2F2%5Ca\".
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\n" ); document.write( " \"x%5B1%5D+=+%28-%286%29%2Bsqrt%28+900+%29%29%2F2%5C3+=+4\"
\n" ); document.write( " \"x%5B2%5D+=+%28-%286%29-sqrt%28+900+%29%29%2F2%5C3+=+-6\"
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\n" ); document.write( " Quadratic expression \"3x%5E2%2B6x%2B-72\" can be factored:
\n" ); document.write( " \"3x%5E2%2B6x%2B-72+=+3%28x-4%29%2A%28x--6%29\"
\n" ); document.write( " Again, the answer is: 4, -6.\n" ); document.write( "Here's your graph:
\n" ); document.write( "\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B6%2Ax%2B-72+%29\"

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\n" ); document.write( "\n" ); document.write( "So we have 2 possible numbers for our first number.
\n" ); document.write( "Now lets check our first possible answer x=4:\r
\n" ); document.write( "\n" ); document.write( "Lets find our 2nd number using x=4:\r
\n" ); document.write( "\n" ); document.write( "\"y=x%2B1\"\r
\n" ); document.write( "\n" ); document.write( "\"y=4%2B1\" Plug in x=4\r
\n" ); document.write( "\n" ); document.write( "\"y=5\"\r
\n" ); document.write( "\n" ); document.write( "So our 2nd number is 5\r
\n" ); document.write( "\n" ); document.write( "Lets find our 3rd number using x=4\r
\n" ); document.write( "\n" ); document.write( "\"z=x%2B2\"\r
\n" ); document.write( "\n" ); document.write( "\"z=4%2B2\" Plug in x=4\r
\n" ); document.write( "\n" ); document.write( "\"z=6\"\r
\n" ); document.write( "\n" ); document.write( "So our 3rd number is 6\r
\n" ); document.write( "\n" ); document.write( "So our 3 numbers are 4,5,6\r
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\n" ); document.write( "\n" ); document.write( "Check:\r
\n" ); document.write( "\n" ); document.write( "\"4%5E2%2B5%5E2%2B6%5E2=77\" Plug in x=4,y=5,z=6\r
\n" ); document.write( "\n" ); document.write( "\"16%2B25%2B36=77\"\r
\n" ); document.write( "\n" ); document.write( "\"77=77\" works\r
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\n" ); document.write( "\n" ); document.write( "Now lets check our other possible answer x=-6:\r
\n" ); document.write( "\n" ); document.write( "Lets find the 2nd number using x=-6\r
\n" ); document.write( "\n" ); document.write( "\"y=x%2B1\"\r
\n" ); document.write( "\n" ); document.write( "\"y=-6%2B1\" Plug in x=-6\r
\n" ); document.write( "\n" ); document.write( "\"y=-5\"\r
\n" ); document.write( "\n" ); document.write( "So our 2nd number is -5\r
\n" ); document.write( "\n" ); document.write( "Lets find our 3rd number using x=-6\r
\n" ); document.write( "\n" ); document.write( "\"z=x%2B2\"\r
\n" ); document.write( "\n" ); document.write( "\"z=-6%2B2\" Plug in x=-6\r
\n" ); document.write( "\n" ); document.write( "\"z=-4\"\r
\n" ); document.write( "\n" ); document.write( "So our 3rd number is -4\r
\n" ); document.write( "\n" ); document.write( "So our 3 numbers are -6,-5,-4\r
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\n" ); document.write( "\n" ); document.write( "Check:\r
\n" ); document.write( "\n" ); document.write( "\"%28-6%29%5E2%2B%28-5%29%5E2%2B%28-4%29%5E2=77\" Plug in x=-6,y=-5,z=-4\r
\n" ); document.write( "\n" ); document.write( "\"36%2B25%2B16=77\"\r
\n" ); document.write( "\n" ); document.write( "\"77=77\" works\r
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\n" ); document.write( "\n" ); document.write( "So we have 2 sets of numbers that work. We have one set\r
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\n" ); document.write( "\n" ); document.write( "4,5,6 \r
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\n" ); document.write( "\n" ); document.write( "and we also have\r
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\n" ); document.write( "\n" ); document.write( "-6,-5,-4\r
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