Question

If AM:GM =13/12 for the two positive numbers then the ratio of the numbers is ​

Answer 1

Step-by-step explanation:

1. step 01 is the 13/12is the question u. change I to 13*12 answer is 156 ok
2. 156*12 is the answer is 1872 ok
3. 1872 is answer ok
4. 12:13 is answer of you question

Answer 2

$\bold{A.M=\dfrac{a+b}{2} > 0}$

$\bold{G.M=\sqrt{ab} > 0}$

Then

$\bold{(A.M)^2=\dfrac{(a+b)^2}{4} > 0}$

$\bold{(G.M)^2=ab > 0}$ ...(1)

$\bold{A.M:G.M =13:12}$

Then

$\bold{(A.M)^2:(G.M)^2 =169:144}$ ...(2)

So from (1), (2)

$\bold{\dfrac{(a+b)^2}{4} :ab=169:144}$

$\bold{\implies (a+b)^2:(4ab)=169:144}$

$\bold{\implies 12^2(a+b)^2=13^2(4ab)}$

$\implies \bold{\dfrac{\cancel{12^2}}{\cancel{2^2}} (a+b)^2=\dfrac{13^2}{\cancel{2^2}}(\cancel{4ab})}$

$\bold{\implies 6^2(a+b)^2=13^2ab}$

$\implies \bold{36(a^2+2ab+b^2)-169ab=0}$

$\bold{\therefore36a^2-97ab+36b^2=0}$

To find the ratio $\bold{a:b=\dfrac{a}{b}}$, divide by ab on both sides.

$\bold{\implies \dfrac{36a^2}{ab} -\dfrac{97ab}{ab} +\dfrac{36b^2}{ab} =0}$

$\bold{\implies \dfrac{36a}{b} -97+\dfrac{36b}{a} =0}$

Substitute $\bold{u=\dfrac{a}{b}}$

$\bold{\implies 36u-117+\dfrac{36}{u} =0}$

$\bold{\implies 36u^2-97u+36=0}$

$\implies\bold{u=\dfrac{97\pm\sqrt{97^2-4\times36} }{72} }$

$\implies\bold{u=\dfrac{97\pm\sqrt{97^2-2^2\times6^2} }{72} }$

$\implies\bold{u=\dfrac{97\pm\sqrt{97^2-12^2} }{72} }$

$\implies\bold{u=\dfrac{97\pm\sqrt{(97+12)(97-12)} }{72} }$

$\implies\bold{u=\dfrac{97\pm\sqrt{(109)(85)} }{72} }$

$\implies\bold{u=\dfrac{97\pm\sqrt{9265} }{72} }$

As u is a ratio, we obtain positive solutions.

But in this form, it is difficult to find signs.

Vieta's formula:

• Real and different solutions
• $\bold{\alpha +\beta =\dfrac{97}{36} >0}$
• $\bold{\alpha \beta =\dfrac{36}{36} =1}$

So, $\bold{\alpha >0}$ and $\bold{\beta >0}$. Both positive.

And we notice that $\bold{\alpha \beta =1}$.

One solution is an inverse of another.

This means the solutions of $\bold{u=\dfrac{a}{b} }$ and $\bold{u=\dfrac{b}{a} }$ will be equal.

Hence, we obtained all solutions.

Therefore, the ratio of the two numbers is $\bold{\dfrac{97\pm\sqrt{9265} }{72} }$∎

Proof that solutions exist:

We know A.M-G.M inequality.

So, A.M is always greater than or equals G.M

$\boxed{\sf{\dfrac{a+b}{2} \geq \sqrt{ab} }}$ ...(A.M-G.M inequality)

We know

$\sf{A.M:G.M=13k:12k}$ for some k>0

$\implies\sf{13k>12k}$, inequality is shown.

$\implies$ Solutions exist.

Learn more about Vieta's formula:

Given a quadratic equation $\bold{ax^2+bx+c=0}$ and two roots are α, β

• $\bold{Sum\:of\:the\:two\:roots=\alpha +\beta =-\dfrac{b}{a} }$
• $\bold{Multiplication\:of\:the\:two\:roots=\alpha \beta =\dfrac{c}{a} }$
• $\bold{Discriminant=b^2-4ac}$

Given that D>0, which is two distinct real solutions

• $\bold{\alpha \beta >0}$ then two equivalent signs
• $\bold{\alpha \beta <0}$ then two different signs
• $\bold{\alpha +\beta >0}$ then two positive solutions (if signs are equal)
• $\bold{\alpha +\beta <0}$ then two negative solutions (if signs are equal)