Question

The number of positive integral solutions of
X^2+9 ‹ (X+3)^2 ‹ 8X+25 is:​

Answer 1

Answer:

first taking the

previous inequality

then

x^2+9<x^2+9+6x

then

6x>0

x>0

now taking last inequality

then

x^2+9+6x<8x+25

then

x^2-2x-16<0

here

roots are

2+√4+64/2

=2+√68/2

=1+√17

and 1-√17

then

{x-(1+√17) }{x-(1-√17) }<0

then

by combining both inequality we get

0<x<1+√17

positive integer is

1,2,3,4,5

ok

follow me

Answer 2

Answer:

None

Step-by-step explanation:

First take 1st part: x^2+9<(x+3)^2

x^2+9-(x+3)^2<0

x^2+9-x^2-6x-9<0

-6x<0

6x>0 [-x sign changed, so inequality sign also changed]

x>0

x belongs to (0,infinty)

Then 2nd part: (x+3)^3<8x+25

x^2+6x+9-(8x+25)<0

x^2-2x-16<0 [There are no real roots]

Hence, there are no positive integral solution.

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