Solve ##\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=4\dfrac{1}{4} \cdots##

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  • #1
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Homework Statement
Solve ##\begin{cases} \sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=4\dfrac{1}{4}\\ \dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}=16\dfrac{1}{4} \end{cases}##
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##\Rightarrow \begin{cases} (\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}})^2=(4\dfrac{1}{4})^2\\ (\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}})^2=(16\dfrac{1}{4})^2 \end{cases}##

##\Leftrightarrow \begin{cases} \dfrac{x}{y}+\dfrac{y}{x}+2=\dfrac{289}{16}\\ \dfrac{x^2}{y}+\dfrac{y^2}{x} +\dfrac{2xy}{\sqrt{xy}}=\dfrac{4225}{16}\end{cases}##

##\Leftrightarrow \begin{cases}\dfrac{x^2+y^2}{xy}=\dfrac{257}{16}\\ \dfrac{x^3+y^3}{xy}+2\sqrt{xy} =\dfrac{4225}{16} \end{cases}##

And then I'm lost here. The presence of ##\sqrt{xy}## in the second equation makes me question if I'm even doing this problem correctly. Did I miss another pattern or simplification in the very first step?
 
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  • #2
If you let ##x = u^2## and ##y = v^2##, then you can eliminate all the square roots.
 
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  • #3
If you let ##z=\sqrt{\dfrac{x}{y}}## you can solve the first equation and then eliminate one unknown from the second equation.
 
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  • #4
In fact, the given rational numbers are such that with the substitution above you don't even need to solve a quadratic equation.
 
  • #5
As you can see my instinct is to just plough ahead with the given equations instead of doing substitutions. Is there any specific characteristics to equations that give a hint to which substitution is most likely to succeed or is this just a case of doing loads of problems and memorizing what type of substitutions work for their corresponding type of equations? Because I'm not yet grasping any intuition with these types of problems. Thanks.
 
  • #6
RChristenk said:
As you can see my instinct is to just plough ahead with the given equations instead of doing substitutions. Is there any specific characteristics to equations that give a hint to which substitution is most likely to succeed or is this just a case of doing loads of problems and memorizing what type of substitutions work for their corresponding type of equations? Thanks.
I inspect the equations looking for patterns and symmetries and then try to utilize them.
 
  • #7
Hill said:
In fact, the given rational numbers are such that with the substitution above you don't even need to solve a quadratic equation.
You must have a very clever solution!
 
  • #8
PeroK said:
You must have a very clever solution!
With the substitution of the post #3, the first equation becomes, $$z+ \frac 1 z = 4 + \frac 1 4$$ The two immediate solutions are ##z=4## and ##z= \frac 1 4##.

Then, ##\sqrt x =4 \sqrt y## (or, interchange ##x## and ##y##), and the second equation becomes as straightforward.
 
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  • #9
Hill said:
With the substitution of the post #3, the first equation becomes, $$z+ \frac 1 z = 4 + \frac 1 4$$ The two immediate solutions are ##z=4## and ##z= \frac 1 4##.

Then, ##\sqrt x =4 \sqrt y## (or, interchange ##x## and ##y##), and the second equation becomes as straightforward.
I had managed to interpret ##4\frac 1 4## as ##\frac 5 4##. That was throwing up complex solutions!
 
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  • #10
RChristenk said:
As you can see my instinct is to just plough ahead with the given equations instead of doing substitutions. Is there any specific characteristics to equations that give a hint to which substitution is most likely to succeed or is this just a case of doing loads of problems and memorizing what type of substitutions work for their corresponding type of equations? Because I'm not yet grasping any intuition with these types of problems. Thanks.
There's no point in doing lots of problems and just hoping things sink in. You have to understand why things work and why things don't. Perhaps you could keep a checklist. For example, an equation of the form:
$$az + \frac c z + b = 0$$ is a quadratic equation in disguise. Try to remember than. Or, have a folder with "useful ideas" and write it down.

Another idea is that the ratio between two things is often useful. In this case ##z = \frac x y## was a good idea.

Another idea is to look at how to get rid of a square root term. Take a variation of that second equation:
$$\sqrt x + \sqrt y = a$$You can't get rid of the square roots in one step, but you can do it in two steps, in various way:
$$x^2 + y^2 + 2\sqrt{xy} = a^2$$$$2\sqrt{xy} = a^2 - x^2 - y^2$$$$4xy = (a^2 - x^2 - y^2)^2$$Finally, you can replace a square root, as I suggested:
$$\sqrt x + \sqrt y = a$$$$u + v = a$$Where ##x = u^2, y = v^2##. That doesn't really change very much, but it makes it easier to work with.

In addition to a sheet of good ideas, have a sheet of "mistakes I must not repeat!". In any case, you have to find some way for these things to sink in.
 
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  • #11
RChristenk said:
Homework Statement: Solve ##\begin{cases} \sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=4\dfrac{1}{4}\\ \dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}=16\dfrac{1}{4} \end{cases}##
Let's generalise this to:
$$\begin{cases} \sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}= a\\ \dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}=b \end{cases}$$If ##a \ge 2## we have real solutions:
$$\sqrt x = \frac{b}{2(a-1)}\bigg [1 + \sqrt{\frac{a-2}{a+2}} \bigg]$$$$\sqrt y = \frac{b}{2(a-1)}\bigg [1 - \sqrt{\frac{a-2}{a+2}} \bigg]$$We can check for ##a = \frac{17}{4}##:
$$\sqrt x, \sqrt y = \frac{2b}{13}\bigg [1 \pm \frac 3 5 \bigg] = \frac{16b}{65}, \frac{4b}{65}$$Moreover, if ##a < 2##, then we have complex solutions:
$$\sqrt x, \sqrt y = \frac{b}{2(a-1)}\bigg [1 \pm i\sqrt{\frac{2-a}{2 +a}} \bigg]$$There's plenty of algebra practice if you want to derive those!
 
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