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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

$${t}={x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5} \\ $$$${t}^{{t}^{{t}} } =\mathrm{2018} \\ $$$$\mathrm{2}^{\mathrm{2}^{\mathrm{2}} } =\mathrm{2}^{\mathrm{4}} =\mathrm{16}<\mathrm{2018} \\ $$$$\mathrm{3}^{\mathrm{3}^{\mathrm{3}} } =\mathrm{3}^{\mathrm{27}} =\mathrm{7625597484987}>\mathrm{2018} \\ $$$$\Rightarrow\mathrm{2}<{t}<\mathrm{3} \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}=\left({x}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{1}\geqslant\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}={t}\:\Rightarrow{real}\:{solution}\:{exists}! \\ $$$$\Sigma{x}=−\mathrm{4} \\ $$

Commented by Tawa1 last updated on 21/Nov/18

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

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{did}\:\mathrm{not}\:\mathrm{really}\:\mathrm{understand}\:\mathrm{sir}. \\ $$

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

$${the}\:{sum}\:{of}\:{the}\:{solutions}\:{of} \\ $$$${x}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${is}\:−{b}. \\ $$$$ \\ $$$${you}\:{can}\:{see}: \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}={t} \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}−{t}=\mathrm{0} \\ $$$$\left({x}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{1}−{t}=\mathrm{0} \\ $$$$\left({x}+\mathrm{2}\right)^{\mathrm{2}} ={t}−\mathrm{1} \\ $$$${x}+\mathrm{2}=\pm\sqrt{{t}−\mathrm{1}} \\ $$$${x}=−\mathrm{2}\pm\sqrt{{t}−\mathrm{1}} \\ $$$${i}.{e}.\: \\ $$$${x}_{\mathrm{1}} =−\mathrm{2}+\sqrt{{t}−\mathrm{1}} \\ $$$${x}_{\mathrm{2}} =−\mathrm{2}−\sqrt{{t}−\mathrm{1}} \\ $$$$\Rightarrow{x}_{\mathrm{1}} +{x}_{\mathrm{2}} =−\mathrm{4} \\ $$

Commented by MJS last updated on 21/Nov/18

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

$$\mathrm{3}^{\mathrm{3}^{\mathrm{3}} } =\mathrm{3}^{\mathrm{27}} =\mathrm{7625597484987}\:\mathrm{but}\:\mathrm{this}\:\mathrm{doesn}'\mathrm{t} \\ $$$$\mathrm{affect}\:\mathrm{your}\:\mathrm{proof} \\ $$

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

thanks for checking sir!

$${thanks}\:{for}\:{checking}\:{sir}! \\ $$

Commented by Tawa1 last updated on 22/Nov/18

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\: \\ $$

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