Question 165493
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Two consecutive ODD Integers ----{{{system(x,x+2)}}}
Condition:
{{{x^2+(x+2)^2=290}}}, EQN 1
Continuing,
{{{x^2+x^2+4x+4=290}}}
{{{2x^2+4x+4-290}}} ---> {{{2x^2+4x-286=0}}}, divide whole eqn by 2
{{{cross(2)x^2/cross(2)+cross(4)2x/cross(2)-cross(289)143/cross(2)=0/2}}}
{{{x^2+2x-143-0}}}
Remember ---{{{system(a=1,b=2,c=-143)}}}
By PYTH.THEOREM:
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}
{{{x=(-2+-sqrt(2^2-4(1)(-143)))/(2*1)}}}
{{{x=(-2+-sqrt(4+572))/2}}}
{{{x=(-2+-sqrt(576))/2}}} ----> {{{x=(-2+-24)/2}}}
2 VALUES:
{{{x=(-2+24)/2=22/2}}}
{{{highlight(x=11)}}}
{{{x=(-2-24)/2=-26/2}}}
{{{x=-13}}}
USE HIGHLIGHTED ONE:
{{{x=11}}} -------------> 1st odd intgr.
{{{x+2=11+2=13}}} ------> 2nd odd intgr.
Check via EQN 1,
{{{11^2+13^2=290}}}
{{{121+169=290}}}
{{{290=290}}}
Thank you,
Jojo</pre>