Find two numbers whose $AM+...$

Question

Find two numbers whose $AM + GM =25$ and $AM:GM=5:3$.

My Attempt;

Given, $\frac {AM}{GM}=\frac {5}{3} = k (let) $ $\AM=5k, GM=3k$.

Also, $AM+GM=25$ $5k+3k=25$ $8k=25$ $k=\frac {25}{8}$.

Am I going right? Or, is there any other simple alternative.?

Answer 1

Your process of working has so far been correct, but there is still work to do.

Substituting we have AM=$\frac {a+b}{2}=\frac {125}{8} $ and GM=$\sqrt {ab}=\frac {75}{8} $ where $a $ and $b $ are the two numbers. Thus, find $(a-b) $ from the relation $$(a-b)=\sqrt {(a+b)^2-4ab} $$ and solve for the numbers. Hope it helps.

Answer 2

we have $$AM=\frac{5}{3}GM$$ from here we get with the first equation: $$\frac{5}{3}GM+GM=25$$ thus we have $$\frac{8}{3}GM=25$$ and $$GM=\frac{75}{8}$$ from here we get $$ab=\left(\frac{75}{8}\right)^2$$ and $$a+b=\frac{375}{12}$$ you can solve one equation e.g. for $a$ and plug these equation in the other one ok with $$b=\frac{375}{12}-a$$ we get $$a\left(\frac{375}{12}-a\right)=\left(\frac{75}{8}\right)^2$$ this is equivalent to $$0=a^2-\frac{375}{12}a+\left(\frac{75}{8}\right)^2$$ solveing this we get $$a=\frac{225}{8}$$ or $$a=\frac{25}{8}$$

Answer 3

Let the numbers be $a^2$ and $b^2$. Then we have : $$ \frac{a^2+b^2}{2} +ab=25 \implies (a+b)^2=50 \implies a+b=5\sqrt{2} $$

Let $AM$ be $5x$ and $GM$ be $3x$. Then, $x=\frac{25}{8}$ and the $AM$ is $\frac{125}{8}$ and the $GM$ is $\frac{75}{8}$. Hence $$ a^2+b^2=2AM=\frac{125}{4} \\ 2ab=2GM=\frac{75}{4} \\ a^2+b^2-2ab=\frac{125-75}{4}\\ a-b=\sqrt{\frac{50}{4}}=\frac{5\sqrt{2}}{2} $$ Find $a$ and $b$ from the two equations and $a^2$ and $b^2$ are your answers.