Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
Square equation $I$. –  Daniel Fischer Aug 26 '13 at 14:19
2  
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

6 Answers 6

up vote 2 down vote accepted

HINT:

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 $3$

share|improve this answer

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

share|improve this answer

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 $x_1,x_2= 4 \pm 1$

share|improve this answer
    
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.

share|improve this answer

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:

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

share|improve this answer

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

so

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.