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14 Questions around this concept.
If p and q are integers and $p^2=q^2+9$, then the number of pairs of $(p, q)$ possible are :-
An irrational equation is an equation where the variable is inside the radical or the variable is a base of power with fractional exponents..
For example,
$
\sqrt{2 x-3}=4
$
To solve any equation given in the form of $\sqrt[n]{f(x)}=g(x)$, where $n \in N$ and $f(x), g(x)$ are polynomial function, solve $f(x)=(g(x))^n$, and check the values of $x$ obtained by putting it in original equation
For example,
given equation is $\sqrt{\mathrm{x}^2-4 \mathrm{x}+4}=\mathrm{x}+1$
then, $x^2-4 x+4=(x+1)^2$
$\Rightarrow \mathrm{x}^2-4 \mathrm{x}+4=\mathrm{x}^2+2 \mathrm{x}+1$
$\Rightarrow 6 \mathrm{x}=3 \Rightarrow \mathrm{x}=\frac{1}{2}$
$\mathrm{x}=\frac{1}{2}$ satisfies the original equation
so $x=1 / 2$ is the answer
Inequalities:
If n is odd
To solve inequations of the form $(\mathrm{f}(\mathrm{x}))^{1 / n}>\mathrm{g}(\mathrm{x})$ or $(\mathrm{f}(\mathrm{x}))^{1 / n}<\mathrm{g}(\mathrm{x})$, or $(\mathrm{f}(\mathrm{x}))^{1 / n}>(\mathrm{g}(\mathrm{x}))^{1 / n}$, raise both sides to the power n , and solve to get the answer.
If n is even
1. To solve inequations of the form $(f(x))^{1 / n}>g(x)$,
a. LHS should be defined, so solve $f(x) \geq 0$
b. Now if $\mathrm{g}(\mathrm{x})<0$, then LHS will be greater than RHS for all such values
c. If $g(x) \geq 0$, then solve $f(x)>(g(x))^n$
In the end, take the intersection of a with (b union c)
2. To solve inequations of the form $(f(x))^{1 / n}<g(x)$,
a. LHS should be defined, so solve $f(x) \geq 0$
b. Now if $\mathrm{g}(\mathrm{x})<0$, then LHS will not be lesser than RHS for all such values
c. If $g(x) \geq 0$, then solve $f(x)<(g(x))^n$
In the end take the intersection of a and c
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