Question

The no. of integral solutions of this inequality:

x²+9 < (x+3)² < 8x+25

Answer 1

(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
Hope it helps
If yes please mark brainliest

Answer 2

there will be infinite number of solution let me explain it.
$three \: inequaities \: will \: form \\ \\ first \\ {x}^{2} + 9 < {x}^{2} + 9 + 6x \\ 0 < 6x \\ here \: x \: belengs \: to \: + integers \\ \\ second \\ {x}^{2} + 9 + 6x < 8x + 25 \\ {x}^{2} - 2x - 16 < 0 \\ here \: discriminant > 0 \\ hence \: for \: every \: real \: values \: it \: will \: be \: less \: than \: 0 \\ \\ third \\ {x}^{2} + 9 < 8x + 25 \\ {x}^{2} - 8x - 16 < 0 \\ again \: disciminant > 0 \: so \: function \: can \: take \: any \: real \: value \: it \: will \: always negative \\ \\ so \: by \: this \: there \: are \: \infty integral \: solutions \: for \: this \: inequality$