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Question-181676




Question Number 181676 by yaslm last updated on 28/Nov/22
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))
defofsquareroot:b=ab=a2a0x7=x7x7=(x7)2x7x7=x14x+49x49x=4x49x=16x49dontexistxRs.t.x7=x7
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
xx7=7..(i)(xx7)(x+x7)=7(x+x7)7(x+x7)=x(x7)x+x7=1(ii)(i)+(ii):2x=8x=16(iii)(i)(ii):2x7=6x7=3(iv)(iii)&(iv):x=16x7=3xR
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
youmusttestyoursolution16167=7169=743=7wrong
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!
ForgotthatIvesquaredbothsides.Ofcoursethesolutionisincorrect!ThanXmuchsir!
Commented by Rasheed.Sindhi last updated on 30/Nov/22
...Had I studied sir floor′s answer  properly!
HadIstudiedsirfloorsanswerproperly!
Commented by Rasheed.Sindhi last updated on 30/Nov/22
Frix sir, please revisit my answers  of this question.
Frixsir,pleaserevisitmyanswersofthisquestion.
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.
SorryIgetnonotificationsfromthisapp,soImabitlateyouranswersarenowcorrect.
Commented by Rasheed.Sindhi last updated on 01/Dec/22
Thanks a lot sir!
Thanksalotsir!
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
xax7b=7ab=7.(i)a2b2=x(x7)=7a2b2ab=77=1a+b=1(ii)(i)+(ii)&(i)(ii)a=4b=3x=4x7=3xR
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
Forx,aRxa=xaSquaringbothsidesx2ax+a2=xa2axa(a+1)=0a=0isasolution2x(a+1)=0x=(a+1)24a+10Insertingthisinthegivenequationgivesa+12a=a12Thisistruefor1a1xa=xacanonlybesolvedfor1a1

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