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I'm struggling with an obviously easy problem:

Find $x,y$:

$I: \; \sqrt x + \sqrt y=8, \quad \quad II: \; \sqrt{xy}=15$

I tried different ways (put $\sqrt x$ from $I$ into $II$) to solve these equations but I always got stuck.

Anyone have a hint for me?

Thanks a lot in advance

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Square equation $I$. – Daniel Fischer Aug 26 '13 at 14:19
Try solving $u+v=8, uv=15$ first – Hagen von Eitzen Aug 26 '13 at 14:22
$\sqrt{xy}=15 \Rightarrow \sqrt{x}=\frac{15}{\sqrt{y}}$ – Vikram Aug 26 '13 at 14:29
up vote 2 down vote accepted


As $a,b$ are the roots of $x^2-(a+b)x+ab=0,$

$\sqrt x,\sqrt y$ are the roots of $t^2-(8)t+15=0$

$\implies t=\frac{8\pm\sqrt{8^2-4\cdot1\cdot15}}2=\frac{8\pm2}2=5$ or $3$

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Hint: Use Vieta formulas. $ $ $ $ $ $

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I think I just got it (sometimes it's already enough to ask the question in puplic to find the answer...)

$II: \; \sqrt{xy}=15 \Leftrightarrow\sqrt x=\frac{15}{\sqrt y}$. Put into $I$:

$\frac{15}{\sqrt y}+\sqrt y=8 \Leftrightarrow15+y=8 \sqrt y \Leftrightarrow y-8 \sqrt y+15=0$

From here I'll get $x_1,x_2= 4 \pm 1$

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When you have squared during the solution, you need to check back in the original equations to make sure you have not introduced a spurious solution. You don't define $x_1,x_2$ but it looks like they are solutions for $\sqrt y$. If so, you are correct. – Ross Millikan Aug 26 '13 at 16:12

If you square the first, the second gets rid of the cross term. Then square the second and the roots are gone.

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For thise, I will use the trick: $$(u - v)^2 = (u+v)^2 - 4uv$$ Substitute $u, v$ by $\sqrt{x}$ and $\sqrt{y}$, we get:

$$\sqrt{x} - \sqrt{y} = \sqrt{ (\sqrt{x} - \sqrt{y})^2 } = \sqrt{(\sqrt{x}+\sqrt{y})^2 - 4\sqrt{xy}} = \sqrt{8^2 - 4\times15} = \pm 2\\ \implies \begin{cases} \sqrt{x} = \frac12 \left((\sqrt{x} + \sqrt{y}) + (\sqrt{x} - \sqrt{y})\right) = \frac12 ( 8 \pm 2 ) & = 4 \pm 1.\\ \sqrt{y} = (\sqrt{x} + \sqrt{y}) - \sqrt{x} = 8 - \sqrt{x} & = 4\mp 1. \end{cases}$$

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No need for Vieta formulas or any tricks, as "by elimination" works. And we also need to justify a step in solving for either root(x) or root(y):

1) root(x) + root(y) = 8

2) root(xy) = 15

From 1) we know x,y >= 0, this justifies root(xy) = root(x)*root(y)

So from 2):

root(x) root(y) = 15

root(y) = 15/root(x)

Sub into 1):

r(x) + 15/r(x) = 8 ... multiply through by r(x)

x + 15 = 8*r(x)

x - 8*r(x) + 15 = 0

We recognize the above is a quadratic in r(x)... the below substitution isn't needed but it helps make the point....

Let u = r(x), thus u^2 = x [And also note, we must have u>=0]

So above eqn becomes:

u^2 - 8u + 15 = 0

(u-5)(u-3) = 0

So u = 5 or 3 ... we can keep both (both nonnegative)

Note: x=u^2 and y = (15/u)^2


u=3 --> x=9, y=25

u=5 --> x=25, y=9

As a final step, CHECK your answers since real number equations dealing with radicals and or steps involving squaring might lead to "extraneous" solutions that need to be thrown away at the end.

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