Question Number 181676 by yaslm last updated on 28/Nov/22
![](https://www.tinkutara.com/question/27984.png)
Answered by floor(10²Eta[1]) last updated on 28/Nov/22
![def of square root: (√b)=a⇔b=a^2 ∧a≥0 (√x)−7=(√(x−7))⇔x−7=((√x)−7)^2 ∧(√x)≥7 ⇔x−7=x−14(√x)+49∧x≥49 ⇔(√x)=4∧x≥49 ⇔x=16∧x≥49 ⇒dont exist x∈R s.t. (√x)−7=(√(x−7))](https://www.tinkutara.com/question/Q181680.png)
Commented by yaslm last updated on 28/Nov/22
great sir
Answered by Rasheed.Sindhi last updated on 30/Nov/22
![(√x) −(√(x−7)) =7........(i) ((√x) −(√(x−7)) )((√x) +(√(x−7)) ) =7((√x) +(√(x−7)) ) 7((√x) +(√(x−7)) )=x−(x−7) (√x) +(√(x−7)) =1.........(ii) (i)+(ii): 2(√x) =8⇒x=16.........(iii) (i)−(ii):−2(√(x−7)) =6 (√(x−7)) =−3......(iv) (iii) & (iv): x=16 ∧(√(x−7)) =−3⇒x∉R](https://www.tinkutara.com/question/Q181728.png)
Commented by Frix last updated on 29/Nov/22
![you must test your solution (√(16))−(√(16−7))=7 (√(16))−(√9)=7 4−3=7 wrong](https://www.tinkutara.com/question/Q181742.png)
Commented by Rasheed.Sindhi last updated on 30/Nov/22
![Forgot that I ′ve squared both sides. ...Of course the solution is incorrect! ThanX much sir!](https://www.tinkutara.com/question/Q181769.png)
Commented by Rasheed.Sindhi last updated on 30/Nov/22
![...Had I studied sir floor′s answer properly!](https://www.tinkutara.com/question/Q181770.png)
Commented by Rasheed.Sindhi last updated on 30/Nov/22
![Frix sir, please revisit my answers of this question.](https://www.tinkutara.com/question/Q181796.png)
Commented by Frix last updated on 30/Nov/22
![Sorry I get no notifications from this app, so I′m a bit late... your answers are now correct.](https://www.tinkutara.com/question/Q181803.png)
Commented by Rasheed.Sindhi last updated on 01/Dec/22
![Thanks a lot sir!](https://www.tinkutara.com/question/Q181823.png)
Answered by Rasheed.Sindhi last updated on 30/Nov/22
![(√x) _(a) −(√(x−7)) _(b) =7⇒a−b=7....(i) a^2 −b^2 =x−(x−7)=7 ((a^2 −b^2 )/(a−b))=(7/7)=1 a+b=1......(ii) (i)+(ii) & (i)−(ii) a=4 ∧ b=−3 (√x)=4 ∧ (√(x−7)) =−3 ⇒x∉R](https://www.tinkutara.com/question/Q181729.png)
Answered by Frix last updated on 30/Nov/22
![For x, a ∈R (√x)−a=(√(x−a)) Squaring both sides x−2a(√x)+a^2 =x−a 2a(√x)−a(a+1)=0 a=0 is a solution 2(√x)−(a+1)=0 x=(((a+1)^2 )/4)∧a+1≥0 Inserting this in the given equation gives ((∣a+1∣)/2)−a=((∣a−1∣)/2) This is true for −1≤a≤1 ⇒ (√x)−a=(√(x−a)) can only be solved for −1≤a≤1](https://www.tinkutara.com/question/Q181806.png)