Question

Determinati multimea solutiilor ecuatiei: 4x^2=1+3+5+7+...+2011+2013+2015

Answer 1

1+3+5+7+...+2011+2013+2015=
=2-1+4-1+6-1+8-1+...+2014-1+2016-1 =
=2(1+2+3+4+...+1007+1008) - 1× 2016:2 =
= 2× 1 008×1 009:2 - 1 008=
=1008×1009-1008=
=1008(1009-1)=
=1008×1008=
=1008^2

4x^2=1+3+5+7+...+2011+2013+2015
=> 4x^2= 1008^2
(2x)^2 = 1008^2
2x= {-1008;1008}
x= {-1008/2;1008/2}
x= {-504;504}

Answer 2

4x^2=1+3+5+7+...+2011+2013+2015
4X^2 = 2×(1+2+......+1007+1008)-2016:2
4X^2 = 2×1009×1008:2-1008
4X^2=1008^2
X=(-504,504)