1.

The no. of integral solutions of this inequality:x²+9 < (x+3)² < 8x+25

Answer»

(x+3)^2 =x^2 +9+6x
This is GREATER than x^2 +9 for every positive value of x
Now CHECKING inequality
(x+3)^2 <8x+25
x^2 - 2x-16<0
This quadratic will show minimum at - b/2a
i.e at x=1
At x=1 value is - 17<0
Thus we have to check only first few positive integers
x=2,-16<0
x=3,-13<0
x=4,-8<0
x=5,-1<0
x=6,8>0(rejected)
There is no need to check after this as we have got minima before this value so after this value will never decrease
Thus
No. Of Integral solutions =5
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