Consider the quadratic equation n x^2+7 sqrt{ | vedclass

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Question

(b)

Given, $n x^2+7 \sqrt{n} x+n=0$ $D=49 n-4 n^2=n(49-4 n)$ $D 
eq 0 ; \quad 	herefore \forall n \in I^{+}$

$	herefore$ Roots are distinct.

Vedclass · Accepted Answer

$I$ and $III$

Vedclass · Answer

(b)

Given, $n x^2+7 \sqrt{n} x+n=0$ $D=49 n-4 n^2=n(49-4 n)$ $D 
eq 0 ; \quad 	herefore \forall n \in I^{+}$

$	herefore$ Roots are distinct.

For roots are real $D \geq 0$

$	herefore \quad n(49-4 n) \geq 0 \Rightarrow n \leq \frac{49}{4}$

So, $n \in\{1,2,3,4, \ldots, 12\}$

So, $x$ have finite value.

Product of roots is $\frac{n}{n}=1$

$	herefore$ Products of root is necessarily integer.

Hence, option $(b)$ is correct.

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