Question 166360
the sum of the squares of six consecutive integers is 1111. what are the integers and what is the problem solving strategy?
<pre><font size = 4 color = "indigo"><b>
Let consecutive the integers be {{{x}}}, {{{x+1}}}, {{{x+2}}}, {{{x+3}}}, {{{x+4}}}, and {{{x+5}}}

Then we have the equation:

{{{x^2+(x+1)^2+(x+2)^2+(x+3)^2+(x+4)^2+(x+5)^2=1111}}}

{{{x^2+(x^2+2x+1)+(x^2+4x+4)+(x^2+6x+9)+(x^2+8x+16)+(x^2+10x+15)=1111}}}

{{{x^2+x^2+2x+1+x^2+4x+4+x^2+6x+9+x^2+8x+16)+x^2+10x+15=1111}}}

Collect terms:

{{{6x^2+30x+55=1111}}}

{{{6x^2+30x-1056=0}}}
 
Divide through by 6:

{{{x^2+5x-176=0}}}

Factor:

{{{(x-11)(x+16)=0}}}

Use the zero factor property:

{{{matrix(2,3,x-11=0,"",x+16=0,x=11,"",x=-16)}}}

So there are two solutions:

If {{{x = 11}}},  then the consecutive integers are

{{{x}}}, {{{x+1}}}, {{{x+2}}}, {{{x+3}}}, {{{x+4}}}, and {{{x+5}}}

or

{{{11}}}, {{{11+1}}}, {{{11+2}}}, {{{11+3}}}, {{{11+4}}}, and {{{11+5}}}

or

{{{11}}}, {{{12}}}, {{{13}}}, {{{14}}}, {{{15}}}, and {{{16}}}

and if {{{x = -16}}},  then the consecutive integers are

{{{x}}}, {{{x+1}}}, {{{x+2}}}, {{{x+3}}}, {{{x+4}}}, and {{{x+5}}}

or

{{{-16}}}, {{{-16+1}}}, {{{-16+2}}}, {{{-16+3}}}, {{{-16+4}}}, and {{{-16+5}}}

or

{{{-16}}}, {{{-15}}}, {{{-14}}}, {{{-13}}}, {{{-12}}}, and {{{-11}}}

Edwin</pre>