Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 183846 by Michaelfaraday last updated on 30/Dec/22

Answered by MJS_new last updated on 31/Dec/22

1. x=0 ⇒ (√(x−(√x)))=0  2. x>0  t=(√(x−(√x)))≥0 ⇒ x=(((1±(√(4t^2 +1)))^2 )/4)  insert in the given equation to get  t=7

$$\mathrm{1}.\:{x}=\mathrm{0}\:\Rightarrow\:\sqrt{{x}−\sqrt{{x}}}=\mathrm{0} \\ $$$$\mathrm{2}.\:{x}>\mathrm{0} \\ $$$${t}=\sqrt{{x}−\sqrt{{x}}}\geqslant\mathrm{0}\:\Rightarrow\:{x}=\frac{\left(\mathrm{1}\pm\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{insert}\:\mathrm{in}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{to}\:\mathrm{get} \\ $$$${t}=\mathrm{7} \\ $$

Commented by Michaelfaraday last updated on 31/Dec/22

okay thank but please give more detail  on it.

$${okay}\:{thank}\:{but}\:{please}\:{give}\:{more}\:{detail} \\ $$$${on}\:{it}. \\ $$

Commented by MJS_new last updated on 31/Dec/22

I give the path you do the writing work. it′s  easy...

$$\mathrm{I}\:\mathrm{give}\:\mathrm{the}\:\mathrm{path}\:\mathrm{you}\:\mathrm{do}\:\mathrm{the}\:\mathrm{writing}\:\mathrm{work}.\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{easy}... \\ $$

Commented by Michaelfaraday last updated on 31/Dec/22

okay sir

$${okay}\:{sir} \\ $$

Answered by mr W last updated on 31/Dec/22

x^2 −50x=49(√x)  let u=(√x) ≥0  u^4 −50u^2 =49u  u=0:  ⇒(√(x−(√x)))=(√(u^2 −u))=0 ✓  u≠0:  ⇒u^3 −50u−49=0  ⇒u^3 +u^2 −u^2 −u−49u−49=0  ⇒(u+1)(u^2 −u−49)=0  ⇒u^2 −u=49   ⇒(√(x−(√x)))=(√(u^2 −u))=(√(49))=7 ✓

$${x}^{\mathrm{2}} −\mathrm{50}{x}=\mathrm{49}\sqrt{{x}} \\ $$$${let}\:{u}=\sqrt{{x}}\:\geqslant\mathrm{0} \\ $$$${u}^{\mathrm{4}} −\mathrm{50}{u}^{\mathrm{2}} =\mathrm{49}{u} \\ $$$${u}=\mathrm{0}: \\ $$$$\Rightarrow\sqrt{{x}−\sqrt{{x}}}=\sqrt{{u}^{\mathrm{2}} −{u}}=\mathrm{0}\:\checkmark \\ $$$${u}\neq\mathrm{0}: \\ $$$$\Rightarrow{u}^{\mathrm{3}} −\mathrm{50}{u}−\mathrm{49}=\mathrm{0} \\ $$$$\Rightarrow{u}^{\mathrm{3}} +{u}^{\mathrm{2}} −{u}^{\mathrm{2}} −{u}−\mathrm{49}{u}−\mathrm{49}=\mathrm{0} \\ $$$$\Rightarrow\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} −{u}−\mathrm{49}\right)=\mathrm{0} \\ $$$$\Rightarrow{u}^{\mathrm{2}} −{u}=\mathrm{49}\: \\ $$$$\Rightarrow\sqrt{{x}−\sqrt{{x}}}=\sqrt{{u}^{\mathrm{2}} −{u}}=\sqrt{\mathrm{49}}=\mathrm{7}\:\checkmark \\ $$

Commented by Michaelfaraday last updated on 31/Dec/22

thanks sir

$${thanks}\:{sir} \\ $$

Answered by manxsol last updated on 31/Dec/22

(√(x−(√x))) =(√((x^2 −x)/(x+(√x))))=7  x^2 −x=49x+49(√x)=49(x+(√x))  ((x^2 −x)/(x+(√x)))=49

$$\sqrt{{x}−\sqrt{{x}}}\:=\sqrt{\frac{{x}^{\mathrm{2}} −{x}}{{x}+\sqrt{{x}}}}=\mathrm{7} \\ $$$${x}^{\mathrm{2}} −{x}=\mathrm{49}{x}+\mathrm{49}\sqrt{{x}}=\mathrm{49}\left({x}+\sqrt{{x}}\right) \\ $$$$\frac{{x}^{\mathrm{2}} −{x}}{{x}+\sqrt{{x}}}=\mathrm{49} \\ $$

Commented by Michaelfaraday last updated on 31/Dec/22

thanks sir

$${thanks}\:{sir} \\ $$

Answered by CElcedricjunior last updated on 31/Dec/22

x^2 −50x−49(√x)=0  posons t=(√x)  =>t^4 −50t^2 −49t=0  =>t=0ou  t^3 −50t−49=0  (t+1)(t^2 −t−49)=0  t=−1 ou t^2 −t−49=0  𝚫=1+196=197  t=0 ou t=−1 ou t=((1∓(√(197)))/2)  t=(√x)=>x=(((1+(√(197)))/2))^2   =>  (√(x−(√x)))=(√((((1+(√(197)))/2))^2 −((1+(√(197)))/2)))                    =(√((198−2+2(√(197))−2(√(197)))/4))                  =(√((196)/4))=(√(49))=7  =>(√(x−(√x)))=7  ========================  ............le celebre cedric junior...........  =====================      (

$$\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{50}\boldsymbol{{x}}−\mathrm{49}\sqrt{\boldsymbol{{x}}}=\mathrm{0} \\ $$$$\boldsymbol{{posons}}\:\boldsymbol{{t}}=\sqrt{\boldsymbol{{x}}} \\ $$$$=>\boldsymbol{{t}}^{\mathrm{4}} −\mathrm{50}\boldsymbol{{t}}^{\mathrm{2}} −\mathrm{49}\boldsymbol{{t}}=\mathrm{0} \\ $$$$=>\boldsymbol{{t}}=\mathrm{0}\boldsymbol{{ou}} \\ $$$$\boldsymbol{{t}}^{\mathrm{3}} −\mathrm{50}\boldsymbol{{t}}−\mathrm{49}=\mathrm{0} \\ $$$$\left(\boldsymbol{{t}}+\mathrm{1}\right)\left(\boldsymbol{{t}}^{\mathrm{2}} −\boldsymbol{{t}}−\mathrm{49}\right)=\mathrm{0} \\ $$$$\boldsymbol{{t}}=−\mathrm{1}\:\boldsymbol{{ou}}\:\boldsymbol{{t}}^{\mathrm{2}} −\boldsymbol{{t}}−\mathrm{49}=\mathrm{0} \\ $$$$\boldsymbol{\Delta}=\mathrm{1}+\mathrm{196}=\mathrm{197} \\ $$$$\boldsymbol{{t}}=\mathrm{0}\:\boldsymbol{{ou}}\:\boldsymbol{{t}}=−\mathrm{1}\:\boldsymbol{{ou}}\:\boldsymbol{{t}}=\frac{\mathrm{1}\mp\sqrt{\mathrm{197}}}{\mathrm{2}} \\ $$$$\boldsymbol{{t}}=\sqrt{\boldsymbol{{x}}}=>\boldsymbol{{x}}=\left(\frac{\mathrm{1}+\sqrt{\mathrm{197}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=> \\ $$$$\sqrt{\boldsymbol{{x}}−\sqrt{\boldsymbol{{x}}}}=\sqrt{\left(\frac{\mathrm{1}+\sqrt{\mathrm{197}}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}+\sqrt{\mathrm{197}}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\frac{\mathrm{198}−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{197}}−\mathrm{2}\sqrt{\mathrm{197}}}{\mathrm{4}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\frac{\mathrm{196}}{\mathrm{4}}}=\sqrt{\mathrm{49}}=\mathrm{7} \\ $$$$=>\sqrt{\boldsymbol{{x}}−\sqrt{\boldsymbol{{x}}}}=\mathrm{7} \\ $$$$======================== \\ $$$$............{le}\:{celebre}\:{cedric}\:{junior}........... \\ $$$$===================== \\ $$$$ \\ $$$$ \\ $$$$\left(\right. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com