Square both sides and arrange you will get Sqrt(20+x) = x^2-20 Let a = sqrt (20+x) So a^2 = 20+x Let b= x^2-20 a-b=0-----eq1 a^2+b=x^2+x ----eq2 eq1+eq2: a^2+a=x^2+x By comparison: a^2=x^2 and a=x So sqrt(20+x)=x 20+x =x^2 x^2-x-20=0 By solving this quadratic equation you will get x= 5 or x=-4 but after verifying , x=5 is the valid solution.
we get , x^4-40x^2-x+380=0 , add 5x^3 , -5x^3 , by fakt. , (x-5)(x^3+5x^2-15x-76)=0 , solu. , x=5 , test , sqrt(20+sqrt(20+5))=sqrt(20+5) , --> x>0 , --> V25=5 , x=5 , same , OK , /// is not part of the basic equation x^3+5x^2-15x-76=0 , (x+4)(x^2+x-19)=0 , x= -4 , x^2+x-19=0 , x=(-1 +/- V77)/2 , x= -4 , (-1+V77)/2 , (-1-V77)/2 , not a solution to the basic task /// ,
Looking at the thumbnail for about half a minute, one concludes that x= 5 is the solution. The uniqueness is more difficult. You see, x cannot be much larger than sqrt(20) >4. let's try 5, es stimmt.
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more simple approach
let 20 + x = y^2 then 20 + y = x^2 and x > √20
substracting, y^2 - x^2 + (y - x) = 0 => (y - x)(y + x + 1) = 0
(case y - x = 0) 20 + x = x^2 => (x - 5)(x - 4) = 0 => x = 5 (x = -4 discarded)
(case y + x + 1 = 0) x^2 = -(x + 1) + 20 => x^2 + x - 19 = 0
x = (-1 ± √77)/2 (since x > √20, both discarded)
Square both sides and arrange you will get
Sqrt(20+x) = x^2-20
Let a = sqrt (20+x)
So a^2 = 20+x
Let b= x^2-20
a-b=0-----eq1
a^2+b=x^2+x ----eq2
eq1+eq2:
a^2+a=x^2+x
By comparison:
a^2=x^2 and a=x
So sqrt(20+x)=x
20+x =x^2
x^2-x-20=0
By solving this quadratic equation you will get x= 5 or x=-4 but after verifying , x=5 is the valid solution.
This is a powerful alternative method. Thanks 👍💯 for your input and support. 🙏🙏🙏
x=(-0.5+0.5Sqrt[77])=(-1+Sqrt[77])/2
-0.5+0.5Sqrt[77]~3.88748219369606103020319415370815478043793841377725179546384781489138232310965314083784657853435287796882549362049529012083432768762587387860733544256135657551151461205014820344532368711712776176803036213530176358415020714866728425630060509305236186000663384649623637666349970666203236511622609556042607924070675399240656605822767316195771544351937244268796910297368157570989041682711680249796176054418217927284352878786865754465672359878928261038577484408497395285153072963173752168281595651295751884866310088361036894620916490334718260310895044274490891209534991607458382400939970216010505771732233074230938952570052430960731094897251686303418127455463928184732956423786504478699242496607346336970427365103995794204369559740351121349649278297724223841868718961144727187194024110059176099150563221377883101165322110984828007470129630272204111508562228477993157311096046318647945291472121503884267851107342247486442355939020256590363971992269971027477759570120022441866224141724714023504766759748534993915617669135331279735810658754652765...
Simplify your answer!
Sqrt[20+Sqrt[20+x]]=x so
x>0 or x=0
[20+Sqrt[20+x]]=x^2 so
x^2-20>0 (or equal)
x> x>4.47214
Sqrt[20+Sqrt[20+x]]=x x=5 It’s in my head.
How do you know it is the only solution?
(-1-Sqrt[77])/2=-0.5-0.5Sqrt[77]
we get , x^4-40x^2-x+380=0 , add 5x^3 , -5x^3 , by fakt. , (x-5)(x^3+5x^2-15x-76)=0 , solu. , x=5 ,
test , sqrt(20+sqrt(20+5))=sqrt(20+5) , --> x>0 , --> V25=5 , x=5 , same , OK ,
/// is not part of the basic equation x^3+5x^2-15x-76=0 , (x+4)(x^2+x-19)=0 , x= -4 , x^2+x-19=0 , x=(-1 +/- V77)/2 ,
x= -4 , (-1+V77)/2 , (-1-V77)/2 , not a solution to the basic task /// ,
Looking at the thumbnail for about half a minute, one concludes that x= 5 is the solution. The uniqueness is more difficult. You see, x cannot be much larger than sqrt(20) >4. let's try 5, es stimmt.