In this question, two equations ($\mathrm{I}$) and ($\mathrm{II}$) are given. You have to solve both the equations and give answer.

$\mathrm{I}. 2p^2 +20p+50=0$

$\mathrm{II}. q^2 =25$

Consider the following inequalities:

$$\begin{gathered} 1. \frac{a^2-b^2}{a^2+b^2}>\frac{a-b}{a+b}, \text{where} a>b>0 \\ 2. \frac{a^3+b^3}{a^2+b^2}>\frac{a^2+b^2}{a+b}, \text{where} a>b>0 \end{gathered}$$

Which of the above is/ are correct?

Two equations are given, you have to solve the both the equations:

$\left(i\right) X^2=196$

$\left(ii\right) Y^2+2y-48=0$

$\mathrm{Quantity} \mathrm{I}=18 {\mathrm{x}}^3{\mathrm{y}}^3, \mathrm{Quantity} \mathrm{II}=12 {\mathrm{x}}^4{\mathrm{y}}^4, \mathrm{if} \mathrm{x}>0 \& \mathrm{y}<0$

Find the relation between these two quantities.

Consider the following statements:
(i) When $0<\mathrm{x}<1$, then $\frac{1}{1+\mathrm{x}}<1-\mathrm{x}+{\mathrm{x}}^2$.

(ii) When $0<\mathrm{x}<1$, then $\frac{1}{1+\mathrm{x}}>1-\mathrm{x}+{\mathrm{x}}^2$.

(iii) When $-1<\mathrm{x}<0$, then $\frac{1}{1+\mathrm{x}}<1-\mathrm{x}+{\mathrm{x}}^2$.

(iv) When $-1<\mathrm{x}<0$, then $\frac{1}{1+\mathrm{x}}>1-\mathrm{x}+{\mathrm{x}}^2$.

Then the correct statements are

I.$(7{)}^{450}>(9{)}^{300}$

II .${\left(2\right)}^{900}>{\left(7\right)}^{450}$