Question

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

Answer 1

`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`