Question 57584
how do i find three consecutive intergers such that the sum of the squares is 77.
Let the 1st integer=x
then the 2nd integer=x+1
then the third integer=x+1+1=x+2
Then the equation to solve is:
{{{x^2+(x+1)^2+(x+2)^2=77}}}
{{{x^2+(x^2+2x+1)+(x^2+4x+4)=77}}}
{{{x^2+x^2+x^2+2x+4x+1+4=77}}}
{{{3x^2+6x+5=77}}}
{{{3x^2+6x+5-77=0}}}
{{{3x^2+6x-72=0}}}
{{{3(x^2+2x-24)=0}}}
{{{(3/3)(x^2+2x-24)=0/3}}}
{{{(x^2+2x-24)=0}}}
{{{(x+6)(x-4)=0}}}
x+6=0 and x-4=0
x+6-6=0-6  and x-4+4=0+4
x=-6 and x=4
There are two possible sets of integers: -6,-5,-4  and 4,5,6
:
Check by seeing if the sum of their squares is = to 77
{{{(-6)^2+(-5)^2+(-4)^2=77}}}
{{{36+25+16=77}}}
77=77 checks
{{{(4)^2+(5)^2+(6)^2=77}}}
{{{16+25+36=77}}}
{{{77=77}}} this checks too.
Happy Calculating!!!