Question

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Answer 1

Answer:

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Answer 2

Given,

$\displaystyle\longrightarrow\sf{ab+bc+ca=0}$

$\displaystyle\longrightarrow\sf{bc+ca=-ab}$

$\displaystyle\longrightarrow\sf{c(a+b)=-ab}$

$\displaystyle\longrightarrow\sf{a+b=-\dfrac{ab}{c}}$

Also,

$\displaystyle\longrightarrow\sf{ab+bc+ca=0}$

$\displaystyle\longrightarrow\sf{ab+ca=-bc}$

$\displaystyle\longrightarrow\sf{a(b+c)=-bc}$

$\displaystyle\longrightarrow\sf{b+c=-\dfrac{bc}{a}}$

Now,

$\displaystyle\longrightarrow\sf{\sqrt{\dfrac{a+b}{b+c}}=\sqrt{\dfrac{\left(-\dfrac{ab}{c}\right)}{\left(-\dfrac{bc}{a}\right)}}}$

$\displaystyle\longrightarrow\sf{\sqrt{\dfrac{a+b}{b+c}}=\sqrt{\dfrac{a^2b}{bc^2}}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\sqrt{\dfrac{a+b}{b+c}}=\dfrac{a}{c}}}}$

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Let $\displaystyle\sf{b=mx}$ and $\displaystyle\sf{c=nx^2}$ for some positive constants $\displaystyle\sf{m}$ and $\displaystyle\sf{n.}$

Then,

$\displaystyle\longrightarrow\sf{a=mx+nx^2}$

Given that a = 16 for x = 2, i.e.,

$\displaystyle\longrightarrow\sf{2m+4n=16\quad\quad\dots(1)}$

And given that a = 2 for x = 1, i.e.,

$\displaystyle\longrightarrow\sf{m+n=2}$

$\displaystyle\longrightarrow\sf{2m+2n=4\quad\quad\dots(2)}$

Subtracting (2) from (1), we get,

$\displaystyle\longrightarrow\sf{2n=12}$

$\displaystyle\longrightarrow\sf{n=6}$

Then,

$\displaystyle\longrightarrow\sf{m=-4}$

So, for x = 5,

$\displaystyle\longrightarrow\sf{a=-4\times5+6\times25}$

$\displaystyle\longrightarrow\sf{\underline{\underline{a=130}}}$