Hi
Since the numerator is 1+x^2, it is always >0.
For the inequality to hold, the LHS should be less than 0 or negative.
This means that the denominator of LHS should be negative now.
=> 2x^2-21x+40<0
Doing middle term split,
2x^2-5x-16x+40= x(2x-5)-8(2x-5)<0
We get two boundaries for x, x=8 and x=2/5
Clearly you can see that the expression will be negative between these 2 values of x,i.e x should lie in (2/5,8)
Hope this helps!
Question : If $\alpha$ and $\beta$ are the roots of equation $x^{2}-x+1=0$, then which equation will have roots $\alpha ^{3}$ and $\beta ^{3}?$
Option 1: $x^{2}+2x+1=0$
Option 2: $x^{2}-2x-1=0$
Option 3: $x^{2}+3x-1=0$
Option 4: $x^{2}-3x+1=0$
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