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anemone
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Suppose that $x$ and $y$ are positive real numbers. Find all real solutions of the equation $\dfrac{2xy}{x+y}+\sqrt{\dfrac{x^2+y^2}{2}}=\sqrt{xy}+\dfrac{x+y}{2}$.
Find real solutions to a system
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In summary, $x$ and $y$ are positive real numbers. Find all real solutions of the equation $\dfrac{2xy}{x+y}+\sqrt{\dfrac{x^2+y^2}{2}}=\sqrt{xy}+\dfrac{x+y}{2}$. I have an idea... Let a = y/x. Two points:1) Watch out for x = 0 later.2) I tacitly multiplied the first fraction by a/a. So watch out of a = 0, as well.Then, after a bit:\dfrac{2a}{a + 1}
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[sp]
As usual with me I can't find an efficient (or in this case even elegant) way to do this. But I had an idea...
\(\displaystyle \dfrac{2xy}{x + y} + \sqrt{ \dfrac{x^2 + y^2}{2} } = \sqrt{xy} + \dfrac{x + y}{2}\)
Let a = y/x.
Two points:
1) Watch out for x = 0 later.
2) I tacitly multiplied the first fraction by a/a. So watch out of a = 0, as well.
Then, after a bit
\(\displaystyle \dfrac{2a}{a + 1} + \dfrac{1}{ \sqrt{2} } ~ \sqrt{a^2 + 1} = \sqrt{a} + \dfrac{1}{2} (a + 1)\)
Isolate the \(\displaystyle \sqrt{a + 1}\) and square. After some more simplifying and clearing the fractions
\(\displaystyle 2 (a^2 + 1)(a + 1)^2 = 16 a^2 + 4 a(a + 1)^2 + (a + 1)^4 - 8 a (a + 1)^2 + ( -16 a(a + 1) + 4 a (a + 1)^3 ) \sqrt{a}\)
Expanding and simplifying
\(\displaystyle 2a^4 + 4a^3 + 4a^2 + 4a + 2 = (a^4 + 14a^2 + 1) + (4a^3 - 4a^2 - 4a + 4) \sqrt{a}\)
\(\displaystyle a^4+ 4a^3 - 10a^2 + 4a + 1 = (4a^3 - 4a^2 - 4a + 4) \sqrt{a}\)
So far my approach for finding a is pretty standard. Isolate one square root, simplify, then the next step would be to isolate the other square root and simplify. But there is an issue with this procedure: If a = 1 (as it will be when we solve it) then we have both sides of this equal to 0! I have chosen to change the variable to \(\displaystyle b^2 = a\). This means we don't have to square out that LHS, which is ugly enough. But again, we have to be careful about values of b now. (Don't worry, it'll all come together.)
So with the substitution and (yet more) simplifying:
\(\displaystyle b^8 - 4 b^7 + 4 b^6 + 4 b^5 - 10 b^4 + 4 b^3 + 4 b^2 - 4 b + 1 = 0\)
The rational root theorem says that \(\displaystyle b = \pm 1\) are the only rational solutions. Instead of some fancy theorems to prove these are the only real solutions, I simply graphed it. It was faster.
So, \(\displaystyle b = \pm 1\) are the only real solutions. Time to work backwards. \(\displaystyle a = b^2 = 1\). Thus \(\displaystyle \dfrac{y}{x} = 1\) and we get that the solutions to the original equation are all y = x. What about the x = 0 thing? y = x = 0 violates the original equation.
So the final solution is all real \(\displaystyle y = x, ~ x \neq 0\).
[/sp]
-Dan
As usual with me I can't find an efficient (or in this case even elegant) way to do this. But I had an idea...
\(\displaystyle \dfrac{2xy}{x + y} + \sqrt{ \dfrac{x^2 + y^2}{2} } = \sqrt{xy} + \dfrac{x + y}{2}\)
Let a = y/x.
Two points:
1) Watch out for x = 0 later.
2) I tacitly multiplied the first fraction by a/a. So watch out of a = 0, as well.
Then, after a bit
\(\displaystyle \dfrac{2a}{a + 1} + \dfrac{1}{ \sqrt{2} } ~ \sqrt{a^2 + 1} = \sqrt{a} + \dfrac{1}{2} (a + 1)\)
Isolate the \(\displaystyle \sqrt{a + 1}\) and square. After some more simplifying and clearing the fractions
\(\displaystyle 2 (a^2 + 1)(a + 1)^2 = 16 a^2 + 4 a(a + 1)^2 + (a + 1)^4 - 8 a (a + 1)^2 + ( -16 a(a + 1) + 4 a (a + 1)^3 ) \sqrt{a}\)
Expanding and simplifying
\(\displaystyle 2a^4 + 4a^3 + 4a^2 + 4a + 2 = (a^4 + 14a^2 + 1) + (4a^3 - 4a^2 - 4a + 4) \sqrt{a}\)
\(\displaystyle a^4+ 4a^3 - 10a^2 + 4a + 1 = (4a^3 - 4a^2 - 4a + 4) \sqrt{a}\)
So far my approach for finding a is pretty standard. Isolate one square root, simplify, then the next step would be to isolate the other square root and simplify. But there is an issue with this procedure: If a = 1 (as it will be when we solve it) then we have both sides of this equal to 0! I have chosen to change the variable to \(\displaystyle b^2 = a\). This means we don't have to square out that LHS, which is ugly enough. But again, we have to be careful about values of b now. (Don't worry, it'll all come together.)
So with the substitution and (yet more) simplifying:
\(\displaystyle b^8 - 4 b^7 + 4 b^6 + 4 b^5 - 10 b^4 + 4 b^3 + 4 b^2 - 4 b + 1 = 0\)
The rational root theorem says that \(\displaystyle b = \pm 1\) are the only rational solutions. Instead of some fancy theorems to prove these are the only real solutions, I simply graphed it. It was faster.
So, \(\displaystyle b = \pm 1\) are the only real solutions. Time to work backwards. \(\displaystyle a = b^2 = 1\). Thus \(\displaystyle \dfrac{y}{x} = 1\) and we get that the solutions to the original equation are all y = x. What about the x = 0 thing? y = x = 0 violates the original equation.
So the final solution is all real \(\displaystyle y = x, ~ x \neq 0\).
[/sp]
-Dan
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- #3
anemone
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Well done, topsquark! You know, I sense your obsession lately with my POTWs and challenge problems, hehehe...I hope so far you found nothing but fun in tackling all those problems!
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[sp]anemone said:
It's not so much that I've been obsessed, it's more that I have actually been able to solve some recently. I usually try to work them out.
[/sp]
-Dan
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MarkFL
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topsquark said:
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[sp](Whispers) Why are we having a conversation using spoilers?[/sp]
-Dan
-Dan
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anemone
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Because it is fun! (Bow)(Emo)(Drink)
1. How do you find real solutions to a system of equations?
To find real solutions to a system of equations, you can use a variety of methods such as substitution, elimination, or graphing. These methods involve manipulating the equations to isolate a variable and then solving for its value.
2. What does it mean to have real solutions to a system of equations?
Having real solutions to a system of equations means that there is at least one set of values that satisfy all of the equations in the system. In other words, these values make all of the equations true when substituted into them.
3. How do you know if a system of equations has no real solutions?
If a system of equations has no real solutions, it means that there is no set of values that can make all of the equations true. This can be determined by graphing the equations and seeing if they intersect at any point. If they do not intersect, then there are no real solutions.
4. Can a system of equations have more than one set of real solutions?
Yes, a system of equations can have more than one set of real solutions. This means that there are multiple sets of values that satisfy all of the equations in the system. These solutions can be found by solving the equations using different methods or by graphing and finding the points of intersection.
5. How can finding real solutions to a system of equations be useful?
Finding real solutions to a system of equations can be useful in a variety of real-world applications. For example, it can be used to solve problems involving multiple variables such as finding the optimal solution to a business or engineering problem. It can also be used to model and predict relationships between different quantities.
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