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Question Number 48295 by Tawa1 last updated on 21/Nov/18

Answered by mr W last updated on 21/Nov/18

t=x^2 +4x+5 t^t^t =2018 2^2^2 =2^4 =16<2018 3^3^3 =3^(27) =7625597484987>2018 ⇒2<t<3 x^2 +4x+5=(x+2)^2 +1≥1 x^2 +4x+5=t ⇒real solution exists! Σx=−4

t=x2+4x+5ttt=2018222=24=16<2018333=327=7625597484987>2018⇒2<t<3x2+4x+5=(x+2)2+1⩾1x2+4x+5=t⇒realsolutionexists!Σx=−4

Commented by Tawa1 last updated on 21/Nov/18

God bless you sir. I did not really understand sir.

Godblessyousir.Ididnotreallyunderstandsir.

Commented by mr W last updated on 21/Nov/18

the sum of the solutions of x^2 +bx+c=0 is −b. you can see: x^2 +4x+5=t x^2 +4x+5−t=0 (x+2)^2 +1−t=0 (x+2)^2 =t−1 x+2=±(√(t−1)) x=−2±(√(t−1)) i.e. x_1 =−2+(√(t−1)) x_2 =−2−(√(t−1)) ⇒x_1 +x_2 =−4

thesumofthesolutionsofx2+bx+c=0is−b.youcansee:x2+4x+5=tx2+4x+5−t=0(x+2)2+1−t=0(x+2)2=t−1x+2=±t−1x=−2±t−1i.e.x1=−2+t−1x2=−2−t−1⇒x1+x2=−4

Commented by MJS last updated on 21/Nov/18

3^3^3 =3^(27) =7625597484987 but this doesn′t affect your proof

333=327=7625597484987butthisdoesn′taffectyourproof

Commented by mr W last updated on 21/Nov/18

thanks for checking sir!

thanksforcheckingsir!

Commented by Tawa1 last updated on 22/Nov/18

God bless you sir.

Godblessyousir.

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