It appears that there may be another way of solving this simply. From the beginning assume sqrt(x) = t and convert the system to 't'. This way we get a reduced cubic t^3 - 8t + 3 = 0, which yields 3 distinct real roots which can be plugged into the expression t + 1/t to get 2 answers.
@@allanmarder456 sqrt(x) = -3 is NOT extraneous. It is a valid solution to the equation, given that it asks for an expression involving sqrt(x). Here is another way to look at it: Let the desired result equal y: sqrt(x) + 1/sqrt(x) = y Let r = sqrt(x) and solve for r: r = y/2 + s/2 * sqrt(y^2 - 4) where s = +1 or -1 Plug r into the given equation r^2 + 3/r = 8 I'll skip the tedious math, but you end up with the equation: (y - 3) * (y^2 + 3*y - 2 + s*(y + 3) * sqrt(y^2 - 4)) = 0 For which one answer is obviously 3. But there is another factor, which looks complicated, but if you manipulate, you eventually find that it only has one solution: y = -10/3 which can also be obtained by plugging sqrt(3) = -3 into the requested expression. So the answer to the video is either 3 or -10/3
@InnocentNeuron : Yup, that is a good way to solve it. I did the same. I was surprised there were two valid answers to the question. Usually when they are phrased like that, there is only a single answer.
@@XJWill1 The solutions to the equation r^2 +3/r -8 =0 where r=sqrt(x) :are approx r=.382 and r =2.618 (obtained by Newton's Iteration). They are also the result of my 2 solutions: 5*(3 + or - sqrt(5)). Also sqrt(x) + 1/sqrt(x) = .382 +(1/.382) = 3 and 2.618 +1/2,618=3 Since these 2 values are conjugates they both lead to the same result ie 3. So you are correct 3 is one of the answers, and -10/3 is also a correct answer if you allow sqrt(x)= -3. I (perhaps mistakenly) thought square roots were assumed positive and discarded that result.
The second method shows that x=9 is an extraneous solution to the resulting cubic. If you plug x=9 into the original equation, you get 9 + 3/sqrt(9) = 9 + 1 = 10 So x cannot be 9. You would have to try the other two values of x (which is difficult to evaluate). EDIT: The other two solutions to the cubic are (7 +- 3sqrt(5))/2. You can find the square roots of these two numbers using a trick. I’ll take (7 + 3sqrt(5))/2 as an example. Assume our square root will be of the form a + b*sqrt(5) where a and b are rational. Then we have a^2 + 2ab*sqrt(5) + 5b^2 = 7/2 + (3/2)sqrt(5) Thus a^2 + 5b^2 = 7/2 and 2ab = 3/2 The second equation gives b = 3/4a. Plug this into the first equation: a^2 + 45/(16a^2) = 7/2 Or (a^2) - (7/2)a^2 + 45/16 = 0 Add 1/4 to both sides to complete the square (49/16 = (7/4)^2) and get (a^2 - 7/4)^2 = 1/4 If a^2 - 7/4 = -1/2, then a^2 = 5/4, which is impossible since a must be rational. So a^2 - 7/4 = 1/2, which gives a^2 = 9/4, or a = +-3/2. These solutions give b = +-1/2. Of course we want the positive solution, so our square root is 3/2 + sqrt(5)/2. You can check that this is, in fact, a square root of (7 + 3sqrt(5))/2
Using x=(x^1/2)^2 and then solving the cubic equation and one of the root x^1/2 = -3 weird but it satisfies the first equation and the second is -10/3. The others are (9+/-5^1/2)/2.
As others have pointed out, your mistake is that x = 9 is NOT a solution to the yellow equation. But more interesting is that the requested expression arguably has two answers, not just one. Either 3 or -10/3. How? It turns out the sqrt(x) = -3 is a solution to the yellow equation: (-3)^2 + 3/(-3) = 9 - 1 = 8 Of course, ordinarily we would say that sqrt(x) = -3 has no solution. But the way this question is phrased, sqrt(x) is just a symbol that is being used to set up the question. We could have called it r instead of sqrt(x), then the question could be rephrased as: r^2 + 3/r = 8 What is r + 1/r ? In that case, using the trick in the video, we have 3/r + 1 = 9 - r^2 (3 + r)/r = (3 - r)*(3 + r) but do NOT divide by (3 + r), instead factor: (3 + r) * (r + 1/r - 3) = 0 So either r = -3 or r + 1/r = 3 If r = -3, then r + 1/r = -10/3 So the answer is either 3 or -10/3
I see as you do that whenever we take the square root of anything we have root1 or root2 selected answers to fit the problem. I also see your (-3)^2 + 3/(-3) = 8 argument when we chose √9 = -3 case and matched up (-3)^2 = 9 argument. Very clever approach to understanding f(x) = √x shouldn't assume a positive case solution always rule I never believed in because Imaginary plane square roots always shows two roots on every square root on the Euler circle for every a + ib.
@@SyberMath... We have three roots! 😂 The root √x = -3, x = 6.85410196624 and x = 0.14589803375 solutions I think we'll see graphically (but √x has to be thought of on it's negative domain also for the graph). 🤔😂
Let sqrt(x)=u We obtain u^2+3/u=8 And u^3-8u+3=0 A solution is u=-3 so... (u+3)(u^2-3u+1)=0 The other solutions are u=(1/2)*{3-sqrt(5)} And u=(1/2)*{3+sqrt(5)} These are the solutions if we assume u as the variable...which means considering sqrt(x). BUT We should remember that x=9 is a perfect square...so..if we consider that sqrt(9)=+3 or -3 we can still consider x=9 as a valid solution but we must take sqrt(9)=-3 in order to satisfy the original equation (the substitution u=sqrt(x) has yielded u=-3)
It's an outstanding problem because it can be solved presuming an impossible solution as a starting point, since we can write the equation as (√x)³ - 8√x + 3 = 0 and suppose that √x = -3 is a root. Then we divide (√x)³ - 8√x + 3 by √x, and get the equation (√x)² - 3√x + 1, which give us √x = (3±√5)/2. Hence the expression √x + 1/√x has two results. An incorrect one, -10/3, if we assume √x = -3 and a right one, 3, if we consider √x = (3±√5)/2. The only way around this conundrum would be redefining the square root of nine as ±3. Even so the proper approach of the second method would never provide such a result demonstrating its falsehood.
Have not seen the video, but change of variable is the solution. Take the first term as a, second one as b, and then you have a system of equations which can easily be solved :)
After the equation is squared, you find x = 9 to be a solution to the squared equation, but it is not a solution to the original equation; 9 + 3/3 = 9 + 1 = 10 is not equal to 8. Let sqrt(x) = u, then u^2 + 3/u = 8 can be written as u^3 - 8 u + 3 = 0, which is true for u = sqrt(x) = - 3. The answer is then: sqrt(x) + 1/sqrt(x) = - 3 + 1/(-3) = - 3 - 1/3 = - 10/3. (In the second method the factor sqrt(x) + 3 is divided away from the equation...) u^3 - 8u + 3 = (u + 3)(u^2 - 3u + 1) = (u + 3)[u - 3/2 + sqrt(5)/2][u - 3/2 - sqrt(5)/2]. The two other solutions, x = [7 +/- 3sqrt(5)]/2 = u^2, are valid though, and both yield the answer: sqrt(x) + 1/sqrt(x) = sqrt[7/2 +/- 3sqrt(5)/2] + 1/sqrt{[7/2 +/- 3sqrt(5)]/2} (= u + 1/u) = [3 +/- sqrt(5)]/2 + 2[3 -/+ sqrt(5)]/{[3 +/- sqrt(5)][3 -/+ sqrt(5)]} = [3 +/- sqrt(5)]/2 + 2[3 -/+ sqrt(5)]/{9 - 5} = [3 +/- sqrt(5)]/2 + [3 -/+ sqrt(5)/2 = 2*3/2 = 3.
@@SyberMath The thing is that this is unacceptable even if you did _not_ require x to be real because, by definition, the (principle) square root √z of a complex number z has a nonnegative real part. That is, √z cannot be −3 for any complex z. By definition, for any real x, the notation √x represents the (unique) nonnegative real number who's square is x, and for any complex z, by definition, the notation √z represents the (unique) complex number with a nonnegative real part who's square is z. Of course, since the square of a real number is nonnegative, √x is undefined within the set of real numbers for any negative real x.
Writing x+, x- for the positive and negative signed versions of x = ((7+-3sqrt(5))/2 it is easy to see that 1/x+ = x-, so that (x+) + (1/x+) = 7 = (x-) + (1/x-). Then clearly (sqrt(x) + 1/sqrt(x))² = 7 + 2 and the result follows. As others have pointed out x = 9 is an extraneous - that is, erroneous - solution intrtroduced when you squared an equation in method 1.
I came up with the same answer using the iterative method. x = (7 + 3 * sqrt(5))/2 This is the only answer that works. All the others are extraneous roots.
As a matter of curiosity is sqrt(x)=-3 a solution to the equation? I was under the impression that by convention sqrt(x) is always considered non-negative. If sqrt(x)=-3 is allowed there are 2 two solutions, If not there is only 1.
If you treat sqrt(x) as a symbol to answer the question, then certainly it can be -3. You could just as easily call it r, and the question ask for r + 1/r. Since the question did not actually ask for x, it is irrelevant whether sqrt(x) = -3 has a solution (convention is that it does not, but a case could be made for x = 9*i^4)
Alternative approach. Let y = \sqrt{x}, and assume that y + (1/y) = k. Then, you have the two equations: 0 = f(y) = y^3 - 8y + 3 and 0 = g(y) = y^2 - ky + 1. At this point, although there is no guarantee that g(y) divides f(y), that is a reasonable first try. polynomial long division then immediately indicates the candidate value of k=3. you can then solve for y, based on g(y) = 0, with k = 3. Then, you can verify this works by checking whether one of the roots of g(y) = 0 satisfies f(y) = 0.
at 5.25...we may go by this x2 - 7x + 1 = 0 x2 + 1 = 7x dividing by x x + 1/x = 7 adding 2 both the sides x + 1/x + 2 = 7 + 2 (rt.x + 1/rt.x)^2 = 9 rt.x + 1/rt.x = 3
sqrt(9)=3 by definition. The square root is always positive by definition. So x=9 is not a valid solution. For example: x^2=9 then x=+3 and x=-3 but sqrt(9)=3 not -3 9+3/sqrt(9)=10 never=8
It could have easily be checked that 9 + 3/3 = 10 and not 8. 😂 Those kinds of problems are something for calculation sado-masos and bring you not an inch further in understanding mathematics.
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It appears that there may be another way of solving this simply. From the beginning assume sqrt(x) = t and convert the system to 't'. This way we get a reduced cubic t^3 - 8t + 3 = 0, which yields 3 distinct real roots which can be plugged into the expression t + 1/t to get 2 answers.
Exactly. I factored the cubic as (t+3)*(t^2 -3t + 1) giving t = sqrt(x)=-3 (extraneous)
and t=sqrt(x)= (3+- sqrt(5))/2
@@allanmarder456 sqrt(x) = -3 is NOT extraneous. It is a valid solution to the equation, given that it asks for an expression involving sqrt(x).
Here is another way to look at it:
Let the desired result equal y:
sqrt(x) + 1/sqrt(x) = y
Let r = sqrt(x) and solve for r:
r = y/2 + s/2 * sqrt(y^2 - 4) where s = +1 or -1
Plug r into the given equation
r^2 + 3/r = 8
I'll skip the tedious math, but you end up with the equation:
(y - 3) * (y^2 + 3*y - 2 + s*(y + 3) * sqrt(y^2 - 4)) = 0
For which one answer is obviously 3. But there is another factor, which looks complicated, but if you manipulate, you eventually find that it only has one solution:
y = -10/3
which can also be obtained by plugging sqrt(3) = -3 into the requested expression.
So the answer to the video is either 3 or -10/3
@InnocentNeuron : Yup, that is a good way to solve it. I did the same. I was surprised there were two valid answers to the question. Usually when they are phrased like that, there is only a single answer.
Which is what I did. This is the best way to find x. But method 2 is better for answering the question!!!
@@XJWill1 The solutions to the equation r^2 +3/r -8 =0 where r=sqrt(x) :are approx r=.382 and r =2.618 (obtained by Newton's Iteration).
They are also the result of my 2 solutions: 5*(3 + or - sqrt(5)). Also sqrt(x) + 1/sqrt(x) = .382 +(1/.382) = 3 and 2.618 +1/2,618=3
Since these 2 values are conjugates they both lead to the same result ie 3. So you are correct 3 is one of the answers, and -10/3 is also a
correct answer if you allow sqrt(x)= -3. I (perhaps mistakenly) thought square roots were assumed positive and discarded that result.
The second method shows that x=9 is an extraneous solution to the resulting cubic. If you plug x=9 into the original equation, you get
9 + 3/sqrt(9) = 9 + 1 = 10
So x cannot be 9. You would have to try the other two values of x (which is difficult to evaluate).
EDIT: The other two solutions to the cubic are (7 +- 3sqrt(5))/2. You can find the square roots of these two numbers using a trick. I’ll take (7 + 3sqrt(5))/2 as an example.
Assume our square root will be of the form a + b*sqrt(5) where a and b are rational. Then we have
a^2 + 2ab*sqrt(5) + 5b^2 = 7/2 + (3/2)sqrt(5)
Thus a^2 + 5b^2 = 7/2 and 2ab = 3/2
The second equation gives b = 3/4a. Plug this into the first equation:
a^2 + 45/(16a^2) = 7/2
Or (a^2) - (7/2)a^2 + 45/16 = 0
Add 1/4 to both sides to complete the square (49/16 = (7/4)^2) and get
(a^2 - 7/4)^2 = 1/4
If a^2 - 7/4 = -1/2, then a^2 = 5/4, which is impossible since a must be rational. So a^2 - 7/4 = 1/2, which gives a^2 = 9/4, or a = +-3/2. These solutions give b = +-1/2. Of course we want the positive solution, so our square root is 3/2 + sqrt(5)/2. You can check that this is, in fact, a square root of (7 + 3sqrt(5))/2
[7+3sqrt(5)]/2 is the fourth power of the golden ratio.
Using x=(x^1/2)^2 and then solving the cubic equation and one of the root x^1/2 = -3 weird but it satisfies the first equation and the second is -10/3. The others are (9+/-5^1/2)/2.
As others have pointed out, your mistake is that x = 9 is NOT a solution to the yellow equation. But more interesting is that the requested expression arguably has two answers, not just one. Either 3 or -10/3. How?
It turns out the sqrt(x) = -3 is a solution to the yellow equation: (-3)^2 + 3/(-3) = 9 - 1 = 8
Of course, ordinarily we would say that sqrt(x) = -3 has no solution. But the way this question is phrased, sqrt(x) is just a symbol that is being used to set up the question. We could have called it r instead of sqrt(x), then the question could be rephrased as:
r^2 + 3/r = 8
What is r + 1/r ?
In that case, using the trick in the video, we have
3/r + 1 = 9 - r^2
(3 + r)/r = (3 - r)*(3 + r) but do NOT divide by (3 + r), instead factor:
(3 + r) * (r + 1/r - 3) = 0
So either r = -3 or r + 1/r = 3
If r = -3, then r + 1/r = -10/3
So the answer is either 3 or -10/3
I think I set x as real, right? In that case, sqrt(x) = -3 would be unacceptable.
@@SyberMath It is irrelevant what x is, since the question did not ask for x.
I see as you do that whenever we take the square root of anything we have root1 or root2 selected answers to fit the problem. I also see your (-3)^2 + 3/(-3) = 8 argument when we chose √9 = -3 case and matched up (-3)^2 = 9 argument. Very clever approach to understanding f(x) = √x shouldn't assume a positive case solution always rule I never believed in because Imaginary plane square roots always shows two roots on every square root on the Euler circle for every a + ib.
@@SyberMath... We have three roots! 😂 The root √x = -3, x = 6.85410196624 and x = 0.14589803375 solutions I think we'll see graphically (but √x has to be thought of on it's negative domain also for the graph). 🤔😂
x =9 is an extraneous solution.
agreed, can't be fitted to the first equation
Hence, you must check potential solutions.
not really because 3/sqrt(x) is greater than 0 so 8-x>0 and x
Let sqrt(x)=u
We obtain u^2+3/u=8
And
u^3-8u+3=0
A solution is u=-3 so...
(u+3)(u^2-3u+1)=0
The other solutions are
u=(1/2)*{3-sqrt(5)}
And
u=(1/2)*{3+sqrt(5)}
These are the solutions if we assume u as the variable...which means considering sqrt(x).
BUT
We should remember that x=9 is a perfect square...so..if we consider that sqrt(9)=+3 or -3 we can still consider x=9 as a valid solution but we must take sqrt(9)=-3 in order to satisfy the original equation (the substitution u=sqrt(x) has yielded u=-3)
It's an outstanding problem because it can be solved presuming an impossible solution as a starting point, since we can write the equation as (√x)³ - 8√x + 3 = 0 and suppose that √x = -3 is a root. Then we divide (√x)³ - 8√x + 3 by √x, and get the equation (√x)² - 3√x + 1, which give us √x = (3±√5)/2.
Hence the expression √x + 1/√x has two results. An incorrect one, -10/3, if we assume √x = -3 and a right one, 3, if we consider √x = (3±√5)/2. The only way around this conundrum would be redefining the square root of nine as ±3. Even so the proper approach of the second method would never provide such a result demonstrating its falsehood.
Have not seen the video, but change of variable is the solution. Take the first term as a, second one as b, and then you have a system of equations which can easily be solved :)
Checking solutions is a good idea since squaring each side (or raising them to any other even power) can introduce extraneous solutions.
After the equation is squared, you find x = 9 to be a solution to the squared equation, but it is not a solution to the original equation; 9 + 3/3 = 9 + 1 = 10 is not equal to 8.
Let sqrt(x) = u, then u^2 + 3/u = 8 can be written as u^3 - 8 u + 3 = 0, which is true for u = sqrt(x) = - 3. The answer is then: sqrt(x) + 1/sqrt(x) = - 3 + 1/(-3) = - 3 - 1/3 = - 10/3.
(In the second method the factor sqrt(x) + 3 is divided away from the equation...)
u^3 - 8u + 3 = (u + 3)(u^2 - 3u + 1) = (u + 3)[u - 3/2 + sqrt(5)/2][u - 3/2 - sqrt(5)/2]. The two other solutions, x = [7 +/- 3sqrt(5)]/2 = u^2, are valid though, and both yield the answer:
sqrt(x) + 1/sqrt(x) = sqrt[7/2 +/- 3sqrt(5)/2] + 1/sqrt{[7/2 +/- 3sqrt(5)]/2} (= u + 1/u) = [3 +/- sqrt(5)]/2 + 2[3 -/+ sqrt(5)]/{[3 +/- sqrt(5)][3 -/+ sqrt(5)]}
= [3 +/- sqrt(5)]/2 + 2[3 -/+ sqrt(5)]/{9 - 5} = [3 +/- sqrt(5)]/2 + [3 -/+ sqrt(5)/2 = 2*3/2 = 3.
I think I set x as real, right? In that case, sqrt(x) = -3 would be unacceptable.
@@SyberMath The thing is that this is unacceptable even if you did _not_ require x to be real because, by definition, the (principle) square root √z of a complex number z has a nonnegative real part. That is, √z cannot be −3 for any complex z. By definition, for any real x, the notation √x represents the (unique) nonnegative real number who's square is x, and for any complex z, by definition, the notation √z represents the (unique) complex number with a nonnegative real part who's square is z. Of course, since the square of a real number is nonnegative, √x is undefined within the set of real numbers for any negative real x.
Writing x+, x- for the positive and negative signed versions of x = ((7+-3sqrt(5))/2 it is easy to see that 1/x+ = x-, so that (x+) + (1/x+) = 7 = (x-) + (1/x-). Then clearly (sqrt(x) + 1/sqrt(x))² = 7 + 2 and the result follows. As others have pointed out x = 9 is an extraneous - that is, erroneous - solution intrtroduced when you squared an equation in method 1.
9 is not a solution to the original equation, as 9 + 3/rt(9) = 10, not 8.
When you squared the equation at 2:44 you added an extraneous solution.
These sorts of problems shouldn't require finding x. There should be a route to the expression in the question. Which is why method 2 is spot on!
Sqrt x= (3+-sqrt5)/2 and the answer is 3 in all cases
Nice!
I came up with the same answer using the iterative method.
x = (7 + 3 * sqrt(5))/2
This is the only answer that works. All the others are extraneous roots.
As a matter of curiosity is sqrt(x)=-3 a solution to the equation? I was under the impression that by convention sqrt(x) is always considered non-negative.
If sqrt(x)=-3 is allowed there are 2 two solutions, If not there is only 1.
If you treat sqrt(x) as a symbol to answer the question, then certainly it can be -3. You could just as easily call it r, and the question ask for r + 1/r. Since the question did not actually ask for x, it is irrelevant whether sqrt(x) = -3 has a solution (convention is that it does not, but a case could be made for x = 9*i^4)
Nice... learn alot...❤️❤️❤️.
Glad to hear that
root x can be -3 and x can be 9 that will satisfy , so root x + 1/rootx = -10/3
the questioned value could only be determined by the factor other than x-9
Alternative approach. Let y = \sqrt{x}, and assume that y + (1/y) = k. Then, you have the two equations: 0 = f(y) = y^3 - 8y + 3 and 0 = g(y) = y^2 - ky + 1. At this point, although there is no guarantee that g(y) divides f(y), that is a reasonable first try. polynomial long division then immediately indicates the candidate value of k=3. you can then solve for y, based on g(y) = 0, with k = 3. Then, you can verify this works by checking whether one of the roots of g(y) = 0 satisfies f(y) = 0.
at 5.25...we may go by this
x2 - 7x + 1 = 0
x2 + 1 = 7x
dividing by x
x + 1/x = 7
adding 2 both the sides
x + 1/x + 2 = 7 + 2
(rt.x + 1/rt.x)^2 = 9
rt.x + 1/rt.x = 3
Nice :)
Thanks 😁
sqrt(9)=3 by definition. The square root is always positive by definition. So x=9 is not a valid solution.
For example: x^2=9 then x=+3 and x=-3 but sqrt(9)=3 not -3
9+3/sqrt(9)=10 never=8
Dai dati con sqrtx=t>0...risuota t^3-8t+3=0...t=-3(no)..t=3+sqrt5/2..t=3-sqrt5/2...per entrambi i t risulta? =3
x = 6.9042
It could have easily be checked that 9 + 3/3 = 10 and not 8. 😂 Those kinds of problems are something for calculation sado-masos and bring you not an inch further in understanding mathematics.