It would be illuminating to explain where the spurious answer arises from the solution scheme presented. In general for such a problem it is helpful to specify if a real solution is sought and (as several others have commented) to look at the domain of the function and also (from simple bounds) where the solution could lie i.e . between zero and +1.
The equation in the announcing picture is different from the one in the solution. It took me some time so solve (x-x^2)^(1/2)+(x^2-x^3)^(1/2) = 1 - and show that there is no real root. :-((
^=read as to the power *=read as square root As per question *{x-(x)^3}+*{x^2-x^3}=1 Let, a^2= x-x^3, b^2= x^2-x^3 So, *(a^2)+*(b^2)=1 a+b=1......eqn1 a=1-b ab=(1-b)×b=b-b^2.......eqn2 According to the law (a-b)^2=(a+b)^2-4ab =1- {4×(b -b^2)} =1-4b+4b^2 =4b^2- 4b +1 =(2b-1)^2 So, *(a-b)^2=*(2b -1)^2 a -b = 2b - 1........eqn3 Eqn1 + eqn3 a+b+a-b=1+2b - 1 2a= 2b a= b Squaring the equation a^2 = b^2 X - x^3 = x^2 - x^3 X = x^2 X -x^2 =0 X(x-1)=0 X=0 & x-1=0 X=0 & x=1
^1/2 implies square root so x-x^3>=0 X€ [-inf,-1] + [0,1] Also x^2-x^3>=0 x^2(1-x)>=0 So X € (-inf,1] from this condition Overall X a=b=1 So we rewrite equation as (X^2+x-1)^2=0 X1=(-1-√5)/2~=-1.608 double root and X2=(-1+√5)/2~=.608 double root Both of then satisfy the domain conditions. Also x^2=1-x identity holds Substituting in original EQ : √(x-x^3)=√(x-x+x^2)=x √(x^2-x^3)=√(x^2-x+x^2)=√(x^2-x+1-x)=1-x So X+1-x=1 true Now this verification leads to some interesting observations. We could replace the original equation by y^2=x-x^3 (1-y)^2=x^2-x^3 y^2-2y+1=x^2-x^3 Subtracting 1-2y=x-x^2 doesn't make solving any easier than before Dividing equations (-1+1/y)^2=X/(1+X) doesn't seem particularly useful
@dan-florinchereches4892 How do you resolve the fact that your solution X1 is not a solution to the original problem? The original expression is clearly greater than 1 for all negative real x < -1. In your use of the "identity" for solutions, X^2=1-X , when substituting back I think you need to beware that sqrt[X^2] is not X for X
It would be illuminating to explain where the spurious answer arises from the solution scheme presented. In general for such a problem it is helpful to specify if a real solution is sought and (as several others have commented) to look at the domain of the function and also (from simple bounds) where the solution could lie i.e . between zero and +1.
Great question, you're right, it's important to analyze the domain and potential solutions for these types of problems! 👏😊
Respected Sir, Good evening
Good evening to you too! 😎Thanks for tuning in! I’m glad you’re here! 🔥
The equation in the announcing picture is different from the one in the solution.
It took me some time so solve (x-x^2)^(1/2)+(x^2-x^3)^(1/2) = 1 - and show that there is no real root.
:-((
It's was an error. Thanks for bringing it to my attention 🙏🙏🙏
^=read as to the power
*=read as square root
As per question
*{x-(x)^3}+*{x^2-x^3}=1
Let,
a^2= x-x^3, b^2= x^2-x^3
So,
*(a^2)+*(b^2)=1
a+b=1......eqn1
a=1-b
ab=(1-b)×b=b-b^2.......eqn2
According to the law
(a-b)^2=(a+b)^2-4ab
=1- {4×(b -b^2)}
=1-4b+4b^2
=4b^2- 4b +1
=(2b-1)^2
So,
*(a-b)^2=*(2b -1)^2
a -b = 2b - 1........eqn3
Eqn1 + eqn3
a+b+a-b=1+2b - 1
2a= 2b
a= b
Squaring the equation
a^2 = b^2
X - x^3 = x^2 - x^3
X = x^2
X -x^2 =0
X(x-1)=0
X=0 & x-1=0
X=0 & x=1
B B B !!!!!!!!
^1/2 implies square root so
x-x^3>=0
X€ [-inf,-1] + [0,1]
Also x^2-x^3>=0 x^2(1-x)>=0
So X € (-inf,1] from this condition
Overall X a=b=1
So we rewrite equation as
(X^2+x-1)^2=0
X1=(-1-√5)/2~=-1.608 double root and
X2=(-1+√5)/2~=.608 double root
Both of then satisfy the domain conditions.
Also x^2=1-x identity holds
Substituting in original EQ :
√(x-x^3)=√(x-x+x^2)=x
√(x^2-x^3)=√(x^2-x+x^2)=√(x^2-x+1-x)=1-x
So X+1-x=1 true
Now this verification leads to some interesting observations. We could replace the original equation by
y^2=x-x^3
(1-y)^2=x^2-x^3 y^2-2y+1=x^2-x^3
Subtracting
1-2y=x-x^2 doesn't make solving any easier than before
Dividing equations
(-1+1/y)^2=X/(1+X) doesn't seem particularly useful
@dan-florinchereches4892 How do you resolve the fact that your solution X1 is not a solution to the original problem? The original expression is clearly greater than 1 for all negative real x < -1.
In your use of the "identity" for solutions, X^2=1-X , when substituting back I think you need to beware that sqrt[X^2] is not X for X
@@tassiedevil2200 nice point. I didn't check that the first radical is increasing from 0 while the second is increasing from √2
B IS NOT PRONOUNCED AS A P !!!!!!!! Are you hard of hearing !!!!???????.
Enunt: (x-x^3) ori (x-x^2). ……????????????in rest ok