Method 1 just proves we should check the domain before starting the algebra. Due to the square root, we concluded x>0 (since x=0 is not a solution). So 2-x is also positive. Therefore the domain is 0 < x < 2. Since x = 4 + 2sqrt(3) > 2, it is discarded.
I had two methods that were both different than what you did. Method 1: divide the first equation (x + 2sqrtx = 2) by sqrtx. You get: sqrtx + 2 = 2/sqrtx double both sides 2sqrtx + 4 = 4/sqrtx use that to substitute into the second equation (X + 4/sqrtx) X + 2sqrtx + 4 then use the first equation to replace x + 2sqrtx with 2 to get 6 Method 2: Notice that x must be positive. if it's zero you get a divide by zero error, if it's negative you are taking the sqrt of a negative make a new variable a such that a > 0 and a^2 = x replace x in the two equations to get: a^2 + 2a = 2 a^2 + 4/a = ? these are easier equations to solve because there is no more square root just solve for a with the quadratic formula since we didn't square both sides, we didn't add extraneous solutions. we get that a = sqrt3 - 1. from there, solving it is the same.
At 3:46 you said rt(4+2rt3) was rt3+1. Numbers usually have 2 square roots, and you ignored the negative one, which is actually the one that gives the correct answer. If you had used the original equation, rtx= (2-x)/2, this would have given rtx=-(1+rt3), then 1/rtx = (1-rt3)/2, and the expression gives 6. If you use the other value of x, i.e. 4-2rt3, you also get 6 by using the same method.
I approached it as follows. From x+2*sqrt(x)=2 we have x=2-2*sqrt(x). Substitute in the expression to evaluate and we get x+4/sqrt(x)=2-2*sqrt(x)+4/sqrt(x). Cross multiplication gives (2*sqrt(x)-2*x+4)/sqrt(x). Substitution of x gives (2*sqrt(x)-2*(2-2*sqrt(x)+4)/sqrt(x). Distribution and simplification of gives 6*sqrt(x)/sqrt(x), which is 6.
To solve this type of problem, it does not always require solving for X. I tried to manipulate the first equation into the second and was able to do so in three steps. Divide both sides of the first equation by sqrt (X) give: sqrt(X)+2=2/sqrt(X). Second step multiply both sides by sqrt(X) - 2. The left side will become: X-4 and the right will become 2-4/sqrt(x). Last step reorganize terms: X+4/sqrt(X)=6.
This is very similar to what I did in my comment here. If we let √x = u to avoid the square roots and to improve readability, then we have u² + 2u = 2 You first divide both sides by √x = u which gives u + 2 = 2u⁻¹ Then you multiply both sides by √x − 2 = u − 2 to get u² − 4 = 2 − 4u⁻¹ and, finally, rearranging terms you find u² + 4u⁻¹ = 6 which is x + 4/√x = 6 What I did was rewrite u² + 2u = 2 as u² + 2u − 2 = 0 and then multiply both sides by u − 2 to get u³ − 6u + 4 = 0 which can be rewitten as u² + 4u⁻¹ = 6. Both methods really boil down to converting a quadratic equation in √x = u into a cubic equation in √x = u by introducing an extra root √x = u = 2. Since the roots of the quadratic are also roots of the cubic a value of √x satisfying x + 2√x = 2 will also satisfy x + 4/√x = 6. Of course the converse is not true because √x = u = 2 is a root of the cubic but not a root of the quadratic. And, indeed, x = 4 satisfies x + 4/√x = 6 but not x + 2√x = 2.
@@NadiehFan I didn’t notice the similarities in our methods until now. All of the solutions that I looked at involved substitutions and/or finding the values of X so I thought my method was unique.
Let us simply divide both sides of the orig. equation by sqrt(x), so receiving x/(sqrt[x]) + 2 = 2/(sqrt[x]). So the fractional part of the searched expression is twice as much, i.e. 4/(sqrt[x]) = 2((sqrt[x] +2). There only remains to add the "x" to this ... i.e. "searched" = x + 2((sqrt[x] +2), from where after distributing we get x + 2(sqrt[x] +4 where the first two members are aqual to 2 (to two 🙂 ), so the result is 6.
Alternative approach: y=\sqrt{x} ==> y^2 + 2y -2 = 0 and y^3 - ky + 4 = 0. However, since (y^2 + 2y - 2) \times (y-2) = y^3 - 6y + 4, you must have k=6.
I’ll try to make this solution easier to understand. Starting with X+2*sqrt(x)=2, divide both sides with sqrt(X) gives sqrt(X)+2=2/sqrt(X). Then multiply both side with sqrt(X)-2. This give us: X-4 on the left side and 2-4/sqrt(X). Rearranging terms yields X+4/sqrt(X)=6
The most straight forward is to write x+2x^0.5+1=3; (t+1)^2=3; t= 3^.5-1 while the second one t= -3^.5-1 does not work. Your "second method" where you play with the constants is far from the mainstream.
I solved x + 2✓x =2 by substituting ✓x by t t^2 + 2t - 2 = 0 ✓x = +-✓3-1 x=4-+2✓3 Substitution of these two values in the expression x+4/✓x gives you the answer 6.
Solve: x + 2sqrt(x) = 2 x + 4/sqrt(x) = ? Begin with factorizing the first equation x + 2sqrt(x) = sqrt(x)(sqrt(x) + 2) = 2 Then divide sqrt(x) on both sides: sqrt(x) + 2 = 2/sqrt(x) - Multiply by two. Now we have that 4/sqrt(x) = 2sqrt(x) + 4 If we substitute this in for the second equation we get x + 2sqrt(x) + 4, but we know the value of x + 2sqrt(x) it’s just 2. Because of that, the entire expression can be reduced to just be 2+4 which is 6. Answer: 6
What I can't understand is the last graph. If we draw y=x+2√x-2 and y=x+4/√x in the XY plane, we can clearly see that the answer is 6 at the intersection point (4,6).
Did you remember that I gave you that problem (I am talking about the last video), thanks a lot, but actualy it was x^(x+1)=10 But thanks in any way!!!! Loves and prayers from Ecuador!!!
I don't think that is solvable with anything other than numerical approximations. To check, I put it through wolfram alpha, and it came back with a numerical value and nothing more. So, seems unlikely that it can be solved.
@@chaosredefined3834 well, I love wolframalfa, but, wolframalfa has some mistakes, because, one day blackpenredpen solve an equation, but wolframalfa didn’t got even a numerical answer!!! But thanks for your replying me!!! Loves and prayers from Ecuador!!!
You did a lot of similar problems before and although I like watching how you solve these it always amazes me how you succeed in making things more difficult than they are. And yes, you even did this exact same problem over a year ago, on September 29th, 2021, although your methods there were different: ruclips.net/video/WLmDWs_HHcU/видео.html Your second method last year was basically as follows. We have x + 2√x = 2 Dividing both sides by √x (which is allowed since x ≠ 0) we get √x + 2 = 2/√x so we have x + 4/√x = x + 2√x + 4 = 2 + 4 = 6 A less straightforward but rather more interesting approach which shows the relationship with cubic equations and Vieta's formulas would be the following. First, to get rid of the square roots, let √x = u so we have u² + 2u = 2 or u² + 2u − 2 = 0 Unlike what you did in your first method a year ago I won't solve this quadratic, but I do want to find the value of x + 4/√x = u² + 4u⁻¹. Let's call this value a, so we have u² + 4u⁻¹ = a or u³ − au + 4 = 0 Now, I am not claiming that this is always possible (it clearly is not), but _suppose_ that we could choose _a_ in such a way that the roots of the quadratic equation u² + 2u − 2 = 0 are also roots of the cubic equation u³ − au + 4 = 0, then √x + 2 = 2 and √x = u will imply u² + 2u = 2 which in turn will imply u² + 4u⁻¹ = a or x + 4/√x = a. In this way we have transformed the problem of finding the value of x + 4/√x _if_ x + 2√x = 2 into the problem of finding the value of _a_ in the equation u³ − au + 4 = 0 _if_ the roots of this equation comprise the roots of the equation u² + 2u − 2 = 0. To solve this problem, we may note that the sum of the roots of u² + 2u − 2 = 0 is −2 whereas the sum of the roots of u³ − au + 4 = 0 is zero since this equation lacks a quadratic term. Accordingly, _if_ it is at all possible to select _a_ in such a way that the roots of the cubic equation comprise the roots of the quadratic equation, then the additional root of the cubic equation can _only_ be u = 2. And, indeed, the product of the two roots of the quadratic equation is −2 whereas the product of the three roots of the cubic equation is −4, consistent with an additional root u = 2. To create a cubic equation which has u = 2 as a root in addition to the roots which are also the roots of the quadratic equation u² + 2u − 2 = 0 we only need to multiply u² + 2u − 2 by u − 2, so the cubic equation we are looking for is (u − 2)(u² + 2u − 2) = 0 or u³ − 6u + 4 = 0 and comparing this with u³ − au + 4 = 0 we can conclude that a = 6. Accordingly, √x + 2 = 2 and √x = u imply u² + 2u = 2 or u² + 2u − 2 = 0 which implies u³ − 6u + 4 = 0 or u² + 4u⁻¹ = 6 or x + 4/√x = 6.
![Kumamoto Masters Japan 2024 | Gregoria Mariska Tunjung (INA) [5] vs. Akane Yamaguchi (JPN) [2] | F](http://i.ytimg.com/vi/EIjZybhsWqA/mqdefault.jpg)
I like to use repeated substitution. To avoid fractional powers put u = sqrt(x)
u^2 + 2u = 2
u^2 = 2 - 2u
? = u^2 + 4/u
= 2 - 2u + 4/u
= 2 - 2u^2/u + 4/u
= 2 - 2(2 - 2u)/u + 4/u
= 2 - 4/u + 4 + 4/u
= 6
Above equation can easily solved
such as x+2√x=2 , devide the equation by √x We get
√x+2=2/√x, multiplied by 2
2√x+4=4/√x
x+ 4/√x=x+2√x+4=2+4=6
Method 1 just proves we should check the domain before starting the algebra. Due to the square root, we concluded x>0 (since x=0 is not a solution). So 2-x is also positive. Therefore the domain is 0 < x < 2. Since x = 4 + 2sqrt(3) > 2, it is discarded.
I had two methods that were both different than what you did.
Method 1: divide the first equation (x + 2sqrtx = 2) by sqrtx. You get:
sqrtx + 2 = 2/sqrtx
double both sides
2sqrtx + 4 = 4/sqrtx
use that to substitute into the second equation (X + 4/sqrtx)
X + 2sqrtx + 4
then use the first equation to replace x + 2sqrtx with 2
to get 6
Method 2:
Notice that x must be positive. if it's zero you get a divide by zero error, if it's negative you are taking the sqrt of a negative
make a new variable a such that a > 0 and a^2 = x
replace x in the two equations to get:
a^2 + 2a = 2
a^2 + 4/a = ?
these are easier equations to solve because there is no more square root
just solve for a with the quadratic formula
since we didn't square both sides, we didn't add extraneous solutions. we get that a = sqrt3 - 1. from there, solving it is the same.
Very nice, mate! :-)
Nice!
At 3:46 you said rt(4+2rt3) was rt3+1. Numbers usually have 2 square roots, and you ignored the negative one, which is actually the one that gives the correct answer. If you had used the original equation, rtx= (2-x)/2, this would have given rtx=-(1+rt3), then 1/rtx = (1-rt3)/2, and the expression gives 6. If you use the other value of x, i.e. 4-2rt3, you also get 6 by using the same method.
I approached it as follows. From x+2*sqrt(x)=2 we have x=2-2*sqrt(x). Substitute in the expression to evaluate and we get x+4/sqrt(x)=2-2*sqrt(x)+4/sqrt(x). Cross multiplication gives (2*sqrt(x)-2*x+4)/sqrt(x). Substitution of x gives (2*sqrt(x)-2*(2-2*sqrt(x)+4)/sqrt(x). Distribution and simplification of gives 6*sqrt(x)/sqrt(x), which is 6.
To solve this type of problem, it does not always require solving for X. I tried to manipulate the first equation into the second and was able to do so in three steps.
Divide both sides of the first equation by sqrt (X) give: sqrt(X)+2=2/sqrt(X).
Second step multiply both sides by sqrt(X) - 2. The left side will become: X-4 and the right will become 2-4/sqrt(x).
Last step reorganize terms:
X+4/sqrt(X)=6.
This is very similar to what I did in my comment here. If we let √x = u to avoid the square roots and to improve readability, then we have
u² + 2u = 2
You first divide both sides by √x = u which gives
u + 2 = 2u⁻¹
Then you multiply both sides by √x − 2 = u − 2 to get
u² − 4 = 2 − 4u⁻¹
and, finally, rearranging terms you find
u² + 4u⁻¹ = 6
which is
x + 4/√x = 6
What I did was rewrite u² + 2u = 2 as u² + 2u − 2 = 0 and then multiply both sides by u − 2 to get u³ − 6u + 4 = 0 which can be rewitten as u² + 4u⁻¹ = 6. Both methods really boil down to converting a quadratic equation in √x = u into a cubic equation in √x = u by introducing an extra root √x = u = 2. Since the roots of the quadratic are also roots of the cubic a value of √x satisfying x + 2√x = 2 will also satisfy x + 4/√x = 6. Of course the converse is not true because √x = u = 2 is a root of the cubic but not a root of the quadratic. And, indeed, x = 4 satisfies x + 4/√x = 6 but not x + 2√x = 2.
@@NadiehFan I didn’t notice the similarities in our methods until now. All of the solutions that I looked at involved substitutions and/or finding the values of X so I thought my method was unique.
Let us simply divide both sides of the orig. equation by sqrt(x), so receiving x/(sqrt[x]) + 2 = 2/(sqrt[x]). So the fractional part of the searched expression is twice as much, i.e. 4/(sqrt[x]) = 2((sqrt[x] +2).
There only remains to add the "x" to this ... i.e. "searched" = x + 2((sqrt[x] +2),
from where after distributing we get
x + 2(sqrt[x] +4
where the first two members are aqual to 2 (to two 🙂 ), so the result is 6.
Sqrt(x)^2 + 2sqrt(x) + 1 = 3 ... then also we can solve and get two values of sqrt(x)
Using the Lambert-Tsallis Wq function: x = 1/(-Wq(-1/sqrt(2)))^2 = 0.5359... with q = 3. So x+4/sqrt(x) = 6.
Alternative approach: y=\sqrt{x} ==> y^2 + 2y -2 = 0 and y^3 - ky + 4 = 0. However, since (y^2 + 2y - 2) \times (y-2) = y^3 - 6y + 4, you must have k=6.
Wow! Cool
Starting with X+2square root X =2. Divide both sides by sq root of X. Then multiply both sides by sq root of X-2. Simplify.
I’ll try to make this solution easier to understand.
Starting with X+2*sqrt(x)=2, divide both sides with sqrt(X) gives sqrt(X)+2=2/sqrt(X). Then multiply both side with sqrt(X)-2. This give us: X-4 on the left side and 2-4/sqrt(X). Rearranging terms yields X+4/sqrt(X)=6
The most straight forward is to write x+2x^0.5+1=3; (t+1)^2=3; t= 3^.5-1 while the second one t= -3^.5-1 does not work. Your "second method" where you play with the constants is far from the mainstream.
I solved x + 2✓x =2 by substituting
✓x by t
t^2 + 2t - 2 = 0
✓x = +-✓3-1
x=4-+2✓3
Substitution of these two values in the expression x+4/✓x gives you the answer 6.
(x)^0,5 cannot be equal -1-(3)^0,5 because of the definition of sqrt that required to be non negative as x
Solve:
x + 2sqrt(x) = 2
x + 4/sqrt(x) = ?
Begin with factorizing the first equation
x + 2sqrt(x) = sqrt(x)(sqrt(x) + 2) = 2
Then divide sqrt(x) on both sides:
sqrt(x) + 2 = 2/sqrt(x) - Multiply by two.
Now we have that 4/sqrt(x) = 2sqrt(x) + 4
If we substitute this in for the second equation we get x + 2sqrt(x) + 4, but we know the value of x + 2sqrt(x) it’s just 2. Because of that, the entire expression can be reduced to just be 2+4 which is 6.
Answer: 6
What I can't understand is the last graph. If we draw y=x+2√x-2 and y=x+4/√x in the XY plane, we can clearly see that the answer is 6 at the intersection point (4,6).
Nice
x + 2 sqrt(x) = 2
x + 4/sqrt(x) = y
First, I don't like radicals in my equations, so rather than square things, let a = sqrt(x)
a^2 + 2a = 2
Very good!
6
Иероглифы
@@mega_mango те иероглифы в мандарине
Did you remember that I gave you that problem (I am talking about the last video), thanks a lot, but actualy it was
x^(x+1)=10
But thanks in any way!!!!
Loves and prayers from Ecuador!!!
I'm sorry but... How solve this problem? I have no ideas...
@@mega_mango either I, but I do some maths for fun
But you can use the w lambert function or the Newton’s method
I don't think that is solvable with anything other than numerical approximations. To check, I put it through wolfram alpha, and it came back with a numerical value and nothing more. So, seems unlikely that it can be solved.
@@chaosredefined3834 well, I love wolframalfa, but, wolframalfa has some mistakes, because, one day blackpenredpen solve an equation, but wolframalfa didn’t got even a numerical answer!!!
But thanks for your replying me!!!
Loves and prayers from Ecuador!!!
Thank you! Likewise! 🙂
This looks like a half parabola
=6
where are you from
TR living in the US
x + 2sqrtx = 2；（sqrtx +1）^2=3 ;;sqrtx= sqrt3 -1 ;; so last=6
You did a lot of similar problems before and although I like watching how you solve these it always amazes me how you succeed in making things more difficult than they are. And yes, you even did this exact same problem over a year ago, on September 29th, 2021, although your methods there were different:
ruclips.net/video/WLmDWs_HHcU/видео.html
Your second method last year was basically as follows. We have
x + 2√x = 2
Dividing both sides by √x (which is allowed since x ≠ 0) we get
√x + 2 = 2/√x
so we have
x + 4/√x = x + 2√x + 4 = 2 + 4 = 6
A less straightforward but rather more interesting approach which shows the relationship with cubic equations and Vieta's formulas would be the following. First, to get rid of the square roots, let √x = u so we have
u² + 2u = 2
or
u² + 2u − 2 = 0
Unlike what you did in your first method a year ago I won't solve this quadratic, but I do want to find the value of x + 4/√x = u² + 4u⁻¹. Let's call this value a, so we have
u² + 4u⁻¹ = a
or
u³ − au + 4 = 0
Now, I am not claiming that this is always possible (it clearly is not), but _suppose_ that we could choose _a_ in such a way that the roots of the quadratic equation u² + 2u − 2 = 0 are also roots of the cubic equation u³ − au + 4 = 0, then √x + 2 = 2 and √x = u will imply u² + 2u = 2 which in turn will imply u² + 4u⁻¹ = a or x + 4/√x = a.
In this way we have transformed the problem of finding the value of x + 4/√x _if_ x + 2√x = 2 into the problem of finding the value of _a_ in the equation u³ − au + 4 = 0 _if_ the roots of this equation comprise the roots of the equation u² + 2u − 2 = 0.
To solve this problem, we may note that the sum of the roots of u² + 2u − 2 = 0 is −2 whereas the sum of the roots of u³ − au + 4 = 0 is zero since this equation lacks a quadratic term. Accordingly, _if_ it is at all possible to select _a_ in such a way that the roots of the cubic equation comprise the roots of the quadratic equation, then the additional root of the cubic equation can _only_ be u = 2. And, indeed, the product of the two roots of the quadratic equation is −2 whereas the product of the three roots of the cubic equation is −4, consistent with an additional root u = 2.
To create a cubic equation which has u = 2 as a root in addition to the roots which are also the roots of the quadratic equation u² + 2u − 2 = 0 we only need to multiply u² + 2u − 2 by u − 2, so the cubic equation we are looking for is
(u − 2)(u² + 2u − 2) = 0
or
u³ − 6u + 4 = 0
and comparing this with
u³ − au + 4 = 0
we can conclude that a = 6. Accordingly, √x + 2 = 2 and √x = u imply u² + 2u = 2 or u² + 2u − 2 = 0 which implies u³ − 6u + 4 = 0 or u² + 4u⁻¹ = 6 or x + 4/√x = 6.