Question

If $x-1.41\sqrt{x}=1$
Then find the value of $x$.

Answer 1

Given that,

$\begin{array}{c}x-1.41\sqrt{x}=1\\ t=\sqrt{x}\\ \Rightarrow t^2=x\\ t^2-1.4t-1=0\\ t^2-\frac{14t}{10}-1=0\\ \Rightarrow \frac{10t^2-14t-10}{10}=0\\ \Rightarrow 10t^2-14t-10=0\\ \Rightarrow 2\left(5t^2-7t-5\right)=0\\ t=\frac{7\pm \sqrt{{\left(-7\right)}^2-4\times 5\times \left(-5\right)}}{2\times 5}\\ t=\frac{7\pm \sqrt{49+100}}{10}\\ \therefore \sqrt{x}=\frac{7\pm \sqrt{149}}{10}\\ \Rightarrow x={\left(\frac{7+\sqrt{149}}{10}\right)}^2\\ \Rightarrow x={\frac{\left(7-\sqrt{149}\right)}{100}}^2\end{array}$