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Question Number 60734 by Tawa1 last updated on 25/May/19

Find the product of the real roots of the equation           (x + 2 + (√(x^2  + 4x + 3)))^5  −  32(x + 2 − (√(x^2  + 4x + 3)))^5   =  31

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{x}\:+\:\mathrm{2}\:+\:\sqrt{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:+\:\mathrm{3}}\right)^{\mathrm{5}} \:−\:\:\mathrm{32}\left(\mathrm{x}\:+\:\mathrm{2}\:−\:\sqrt{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:+\:\mathrm{3}}\right)^{\mathrm{5}} \:\:=\:\:\mathrm{31} \\ $$

Commented by Prithwish sen last updated on 25/May/19

∵ (x+2+(√(x^2 +4x+3))) = (1/((x+2 −(√(x^2 +4x+3)))))  Now let   (x+2+(√(x^2 +4x+3))) = a  then the equation  a^5 −((32)/a^5 ) =31  ⇒(a^5 +1) −32(((a^5 +1)/a^5 )) =0  either              or,  a^5 =−1               ( (2/a))^5 =1  is it ok ?

$$\because\:\left(\mathrm{x}+\mathrm{2}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{3}}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{2}\:−\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{3}}\right)} \\ $$$$\mathrm{Now}\:\mathrm{let}\: \\ $$$$\left(\mathrm{x}+\mathrm{2}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{3}}\right)\:=\:\mathrm{a} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{a}^{\mathrm{5}} −\frac{\mathrm{32}}{\mathrm{a}^{\mathrm{5}} }\:=\mathrm{31} \\ $$$$\Rightarrow\left(\mathrm{a}^{\mathrm{5}} +\mathrm{1}\right)\:−\mathrm{32}\left(\frac{\mathrm{a}^{\mathrm{5}} +\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }\right)\:=\mathrm{0} \\ $$$$\mathrm{either}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{or}, \\ $$$$\mathrm{a}^{\mathrm{5}} =−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\:\frac{\mathrm{2}}{\mathrm{a}}\right)^{\mathrm{5}} =\mathrm{1} \\ $$$$\mathrm{is}\:\mathrm{it}\:\mathrm{ok}\:? \\ $$

Commented by Tawa1 last updated on 25/May/19

God bless you sir

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

Commented by MJS last updated on 25/May/19

great!  my thinking is too complicated sometimes

$$\mathrm{great}! \\ $$$$\mathrm{my}\:\mathrm{thinking}\:\mathrm{is}\:\mathrm{too}\:\mathrm{complicated}\:\mathrm{sometimes} \\ $$

Commented by Prithwish sen last updated on 25/May/19

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by MJS last updated on 25/May/19

(a+b)^5 −32(a−b)^5 =31  T_1 −32T_2 =31 ⇒ T_1 =31+32T_2   trying some easy numbers I found these:  (a+b)=−1∧(a−b)=−1  x+2±(√(x^2 +4x+3))=−1  (√(x^2 +4x+3))=±(x+3)  (√((x+3)(x+1)))=±(x+3)  (x+3)(x+1)=(x+3)^2   ⇒ x=−3 in both cases    (a+b)^5 =32∧(a−b)^5 =(1/(32))  x+2+(√(x^2 +4x+3))=((32))^(1/5)      x+2−(√(x^2 +4x+3))=(1/((32))^(1/5) )  (√(x^2 +4x+3))=−x      (√(x^2 +4x+3))=x+(3/2)  x^2 +4x+3=x^2             x^2 +4x+3=(x+(3/2))^2   x=−(3/4)                         x=−(3/4)    of course we don′t yet know if there are more  solutions

$$\left({a}+{b}\right)^{\mathrm{5}} −\mathrm{32}\left({a}−{b}\right)^{\mathrm{5}} =\mathrm{31} \\ $$$${T}_{\mathrm{1}} −\mathrm{32}{T}_{\mathrm{2}} =\mathrm{31}\:\Rightarrow\:{T}_{\mathrm{1}} =\mathrm{31}+\mathrm{32}{T}_{\mathrm{2}} \\ $$$$\mathrm{trying}\:\mathrm{some}\:\mathrm{easy}\:\mathrm{numbers}\:\mathrm{I}\:\mathrm{found}\:\mathrm{these}: \\ $$$$\left({a}+{b}\right)=−\mathrm{1}\wedge\left({a}−{b}\right)=−\mathrm{1} \\ $$$${x}+\mathrm{2}\pm\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}}=−\mathrm{1} \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}}=\pm\left({x}+\mathrm{3}\right) \\ $$$$\sqrt{\left({x}+\mathrm{3}\right)\left({x}+\mathrm{1}\right)}=\pm\left({x}+\mathrm{3}\right) \\ $$$$\left({x}+\mathrm{3}\right)\left({x}+\mathrm{1}\right)=\left({x}+\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{x}=−\mathrm{3}\:\mathrm{in}\:\mathrm{both}\:\mathrm{cases} \\ $$$$ \\ $$$$\left({a}+{b}\right)^{\mathrm{5}} =\mathrm{32}\wedge\left({a}−{b}\right)^{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{32}} \\ $$$${x}+\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}}=\sqrt[{\mathrm{5}}]{\mathrm{32}}\:\:\:\:\:{x}+\mathrm{2}−\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}}=\frac{\mathrm{1}}{\sqrt[{\mathrm{5}}]{\mathrm{32}}} \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}}=−{x}\:\:\:\:\:\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}}={x}+\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}={x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}=\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${x}=−\frac{\mathrm{3}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$$$\mathrm{of}\:\mathrm{course}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{yet}\:\mathrm{know}\:\mathrm{if}\:\mathrm{there}\:\mathrm{are}\:\mathrm{more} \\ $$$$\mathrm{solutions} \\ $$

Commented by Tawa1 last updated on 25/May/19

God bless you sir

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

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