If sqrt2 lies between (x+3)/xand (x+4)/(x+1), find the integral value of x ?

1 Answer
Jul 22, 2017

x=7

Explanation:

If x>0, we have x(x+4)=x^2+4x and (x+3)(x+1)=x^2+4x+3, we have

(x+3)(x+1) > x(x+4)

i.e. (x+4)/(x+1) < sqrt2 < (x+3)/x

Hence x+4 < sqrt2x+sqrt2 i.e. x(sqrt2-1)>4-sqrt2

or x>(4-sqrt2)/(sqrt2-1)

and as (4-sqrt2)/(sqrt2-1)=(4-sqrt2)/(sqrt2-1)xx(sqrt2+1)/(sqrt2+1)

= 4-2+3sqrt2=2+4.2426=6.2426 i.e. x>6.2426 (A)

Further, as (x+3)/x > sqrt2,

(sqrt2-1)x <3 i.e. x < 3/(sqrt2-1)

or x < 3sqrt2+3=7.2426 (B)

As such from (A) and (B), 6.2426 < x < 7.2426 and as x is an integer x=7