91Ó°ÊÓ

The given equation is $10x − 17\sqrt{x} + 3 = 0$.

Solving $10x − 17\sqrt{x} + 3 = 0$

$10x+3=17\sqrt{x}$

Squaring both sides , we get:-

$$\begin{gathered} 289x=100x^2+60x+9 \\ 100x^2+60x+9-289x=0 \\ 100x^2-229x+9=0 \end{gathered}$$

Using Quadratic formula:-

$$\begin{gathered} x_{1, 2}=\frac{-\left(-229\right)Â\pm \sqrt{{\left(-229\right)}^2-4· 100· 9}}{2· 100} \\ {x}_{1, 2}=\frac{-\left(-229\right)Â\pm  221}{2· 100} \\ {x}_1=\frac{-\left(-229\right)+221}{2· 100}, x_2=\frac{-\left(-229\right)-221}{2· 100} \\ x=\frac{9}{4}, x=\frac{1}{25} \end{gathered}$$

Plug the value of $x=\frac{9}{4}$in equation $10x − 17\sqrt{x} + 3 = 0$, we get:-

$\begin{array}{rcl}10\left(\frac{9}{4}\right)-17\sqrt{\frac{9}{4}}+3& =& \frac{45}{2}-17\left(\frac{3}{2}\right)+3\\ \frac{45}{2}-17\left(\frac{3}{2}\right)+3& =& -\frac{6}{2}+3\\ -3+3& =& 0\end{array}$

Similarly , Plug $x=\frac{1}{25}$in equation $10x − 17\sqrt{x} + 3 = 0$, we get:-

$\begin{array}{rcl}10\left(\frac{1}{25}\right) − 17\sqrt{\frac{1}{25}} + 3 & =&  \frac{2}{5}-\frac{17}{5}+3\\ \frac{2}{5}-\frac{17}{5}+3& =& \frac{2-17}{5}+3\\ \frac{2-17}{5}+3& =& -3+3\\ -3+3& =& 0\end{array}$

Since, For the value $x=\frac{9}{4}, x=\frac{1}{25}$the equation satisfied the condition. Hence, The solution of the equation is $x=\frac{9}{4}, x=\frac{1}{25}$.