Find the range of values of $x$ for which

${\displaystyle \sqrt{x}+\frac{1}{\sqrt{x}}<4.}$

where $\sqrt{x}$ is the positive root.

The following solution is from Ang Zhi Ping, River Valley High School, Singapore. Hyeyoun Chung, St Paul's Girls' School, London and Yatir Halevi from Maccabim and Reut High School, Israel also sent excellent solutions.

Taking $\sqrt{(}x)=p$, thus we are solving for $p+1/p<4$.

Multiplying both sides by the positive number $p$ a quadratic inequality is obtained, namely $p^2+1<4p\hspace{0.5em}$ or $p^2-4p+1<0$. To factorise the quadratic expression, we find the roots by using the formula

${\displaystyle p=\frac{-b\pm \sqrt{(}{b}^2-4\mathrm{ac})}{2a},}$

where a, b and c are the coefficients of $p^2$, $p$ and the constant in the quadratic expansion. The given inequality holds when

${\displaystyle (p-(2+\sqrt{3}))(p-(2-\sqrt{3}))<0.}$

Knowing that $(p-(2+\sqrt{(}3))<(p-(2-\sqrt{(}3))$ for all real values of p, to make the product negative for all values of $p$ it follows that $(p-(2+\sqrt{(}3))$ must be the negative factor and $(p-(2-\sqrt{(}3))$ the positive factor.

So, the intersection of both ranges $p-(2+\sqrt{3})<0$ and $p-(2-\sqrt{3})>0$ is found to be

${\displaystyle 2-\sqrt{3}<p<2+\sqrt{3}.}$

Substituting $p=\sqrt{x}$, we eliminate the square root by squaring the whole inequality, thus, we get the answer as:

${\displaystyle 7-4\sqrt{3}<x<7+4\sqrt{3}.}$