# Find the value of a?

Let $a$ and $b$ be two positive numbers such that $a\gt b$. Let $G$ be the geometric mean of $a$ and $b$ (that is, $G=\sqrt{ab}$), and $H$ be the Harmonic mean of $a$ and $b$, that is, $$H = \frac{2}{\frac{1}{a}+\frac{1}{b}} = \frac{2ab}{a+b}.$$

If $4G = 5H$, what is the value of $a$?

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Not sure exactly what you're looking for, but $4a^2+4b^2=25ab$ could be considered a quadratic in $a$, so you could solve for $a$ that way in terms of $b$. – yunone Feb 24 '12 at 5:01
I think the OP hasn't put in efforts in solving. Someone please close the question. – Abhishek Parab Feb 24 '12 at 5:02
@AbhishekParab: Closing the question 10 minutes after it has been posted by a relatively new user (2 days in) is rather harsh. Explaining what to do to improve the question is a rather better course. – Arturo Magidin Feb 24 '12 at 5:04
@Abhishek Parab: You don't know that and we don't close questions for that reason anyway. // anna: The equation should entail $4(a+b)^2=25ab$, which is slightly different from what you have. – anon Feb 24 '12 at 5:06
@Ross: I think you are linking to the wrong question... your link is this question. – Arturo Magidin Feb 24 '12 at 5:37

The Harmonic Mean of $a$ and $b$ is $$\frac{2}{\frac{1}{a}+\frac{1}{b}} = \frac{2ab}{a+b}.$$ The Geometric Mean of $a$ and $b$ is $$\sqrt{ab}.$$ So, to state the problem you have in a way that would be actually intelligible would be:

Let $a$ and $b$ be positive numbers such that $a\gt b$; assume that $$4\times\text{geometric-mean(a,b)} = 4\sqrt{ab} = 5\left(\frac{2ab}{a+b}\right) = 5\times\text{harmonic-mean}(a,b).$$ What is the value of $a$?

We have \begin{align*} 4\sqrt{ab} &= \frac{10ab}{a+b}\\ 4(a+b) &= \frac{10ab}{\sqrt{ab}}\\ 2(a+b) &= 5\sqrt{ab}\\ 4(a+b)^2 &= 25ab\\ 4a^2 + 8ab + 4b^2 &= 25ab\\ 4a^2 -17ab + 4b^2 &=0. \end{align*} You can view this as a quadratic equation in $a$; the solutions are given by $$\frac{17b - \sqrt{(17b)^2 - 64b^2}}{8} = \frac{17b-\sqrt{225b^2}}{8} = \frac{17b-15b}{8} = \frac{b}{4}$$ (which is impossible since $a\gt b$) and $$\frac{17b + \sqrt{(17b)^2 - 64b^2}}{8} = \frac{17b + \sqrt{225b^2}}{8} = \frac{32b}{8} = 4b.$$ So the answer is that $a$ must be $4b$.

You can verify this works: the Geometric Mean of $b$ and $4b$ is $\sqrt{4b^2} = 2b$; the Harmonic mean is $$\frac{2(4b)b}{4b+b} = \frac{8b^2}{5b} = \frac{8b}{5}.$$ And $$4(2b) = 5\left(\frac{8b}{5}\right).$$

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Thanks Arturo. I did not think of getting the roots using Discriminant. thank you..:) – vikiiii Feb 24 '12 at 5:20
@anna: You also did your algebra incorrectly, since you went from $2(a+b) = 5\sqrt{ab}$ to $4(a^2+b^2)=25ab$; that's wrong, because $(a+b)^2\neq a^2+b^2$. – Arturo Magidin Feb 24 '12 at 5:21

According to the given information we have, $$4\sqrt{ab} = 5(\frac{2ab}{a+b})$$

$$(a+b) =\frac{5}{2}\sqrt{ab})$$

$$\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}} = \frac{5}{2}$$

Let , $t$ $=$ $\sqrt{\frac{a}{b}}$

$$[t+\frac{1}{t}=\frac{5}{2}].....................Eq(1)$$

A clever person will immediately infer that $t=\frac{1}{2}$

But if its a subjective question we have to justify that also, so

$$({t+\frac{1}{t}})^2=\frac{25}{4}$$

$$t^2 +\frac{1}{t^2} = \frac{17}{4}$$ NOW, $$(t-\frac{1}{t})^2= t^2+\frac{1}{t^2} -2 =\frac{9}{4}$$

$$[t-\frac{1}{t}=\frac{3}{2} ] .....................Eq(2)$$ neglecting the negative value as we know that$L.H.S.>0$ , since $,t>0$

From Eq(1) and Eq(2) we have $t=4$, hence $a=4b$

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