If $\sqrt{2}$ $sqrt2$ lies between $\frac{x + 3}{x}$ $(x+3)/x$ and $\frac{x + 4}{x + 1}$ $(x+4)/(x+1)$ , find the integral value of $x$ $x$ ?

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If $\sqrt{2}$ $sqrt2$ lies between $\frac{x + 3}{x}$ $(x+3)/x$ and $\frac{x + 4}{x + 1}$ $(x+4)/(x+1)$ , find the integral value of $x$ $x$ ?

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Answer 1 · 2017-07-22T14:28:00

$x = 7$ $x=7$

Explanation:

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

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

i.e. $\frac{x + 4}{x + 1} < \sqrt{2} < \frac{x + 3}{x}$ $(x+4)/(x+1) < sqrt2 < (x+3)/x$

Hence $x + 4 < \sqrt{2} x + \sqrt{2}$ $x+4 < sqrt2x+sqrt2$ i.e. $x (\sqrt{2} - 1) > 4 - \sqrt{2}$ $x(sqrt2-1)>4-sqrt2$

or $x > \frac{4 - \sqrt{2}}{\sqrt{2} - 1}$ $x>(4-sqrt2)/(sqrt2-1)$

and as $\frac{4 - \sqrt{2}}{\sqrt{2} - 1} = \frac{4 - \sqrt{2}}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1}$ $(4-sqrt2)/(sqrt2-1)=(4-sqrt2)/(sqrt2-1)xx(sqrt2+1)/(sqrt2+1)$

= $4 - 2 + 3 \sqrt{2} = 2 + 4.2426 = 6.2426$ $4-2+3sqrt2=2+4.2426=6.2426$ i.e. $x > 6.2426$ $x>6.2426$ (A)

Further, as $\frac{x + 3}{x} > \sqrt{2}$ $(x+3)/x > sqrt2$ ,

$(\sqrt{2} - 1) x < 3$ $(sqrt2-1)x <3$ i.e. $x < \frac{3}{\sqrt{2} - 1}$ $x < 3/(sqrt2-1)$

or $x < 3 \sqrt{2} + 3 = 7.2426$ $x < 3sqrt2+3=7.2426$ (B)

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

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If $\sqrt{2}$ $sqrt2$ lies between $\frac{x + 3}{x}$ $(x+3)/x$ and $\frac{x + 4}{x + 1}$ $(x+4)/(x+1)$ , find the integral value of $x$ $x$ ?

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Explanation:

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If $\sqrt{2}$ $sqrt2$ lies between $\frac{x + 3}{x}$ $(x+3)/x$ and $\frac{x + 4}{x + 1}$ $(x+4)/(x+1)$ , find the integral value of $x$ $x$ ?