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Solve for $x \in \mathbb{R}$

${(6x + 5)^2} = \frac{{35}}{{3{x^2} + 5x + 2}}$

We have, $36{x^2} + 60x + 25 = \frac{{35}}{{3{x^2} + 5x + 2}}$

$ \Rightarrow 12(3{x^2} + 5x) + 25 = \frac{{35}}{{(3{x^2} + 5x) + 2}}$

Let, $3{x^2} + 5x = t$

So, $12t + 25 = \frac{{35}}{{t + 2}}$

$ \Rightarrow 12{t^2} + 49t + 15 = 0$

$ \Rightarrow t = \frac{{ - 49 \pm \sqrt {{{49}^2} - 720} }}{{24}} = \frac{{ - 49 \pm 41}}{{24}} =  - \frac{1}{3}, - \frac{{15}}{4}$

Now, $3{x^2} + 5x =  - \frac{1}{3}, - \frac{{15}}{4}$

$9{x^2} + 15x + 1 = 0$ equation yields, $x = \frac{{ - 5 \pm \sqrt {21} }}{6}$

The other equation $12{x^2} + 20x + 15 = 0$ has discriminant < 0, hence no real solutions.

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$f(x)=x^6+2x^4+x^3+2x+3 $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$

$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$