91Ó°ÊÓ

The given equation is $12x + 5\sqrt{x} − 3 = 0$.

Solving $12x + 5\sqrt{x} − 3 = 0$

$$\begin{gathered} {\left(12x− 3\right)}^2={\left(-5\sqrt{x}\right)}^2  \\ 144x^2-72x+9=25x \\ 144x^2-72x-25x+9=0 \\ 144x^2-97x+9=0 \end{gathered}$$

Solving $144x^2-97x+9=0$

$\begin{array}{rcl}{x}_{1, 2}& =& \frac{-\left(-97\right)Â\pm \sqrt{{\left(-97\right)}^2-4· 144· 9}}{2· 144}\\ x_{1, 2}& =& \frac{-\left(-97\right)Â\pm  65}{2· 144}\\ x_1& =& \frac{-\left(-97\right)+65}{2· 144}, x_2=\frac{-\left(-97\right)-65}{2· 144}\\ x& =& \frac{9}{16}, x=\frac{1}{9}\end{array}$

Plug the value of $x=\frac{9}{16}$in equation $12x + 5\sqrt{x} − 3 = 0$, we get:-

$\begin{array}{rcl}12\left(\frac{9}{16}\right) + 5\sqrt{\frac{9}{16}} − 3 & =&  \frac{27}{4}+\frac{15}{4}-3\\ \frac{27}{4}+\frac{15}{4}-3& =& \frac{42}{4}-3\\ \frac{42}{4}-3& =& \frac{30}{4}\\ \frac{30}{4}& =& \frac{15}{2}\end{array}$

Since the value of $\frac{9}{16}$is the equation satisfied the condition.

Similarly, Plug value of $x=\frac{1}{9}$in equation $12x + 5\sqrt{x} − 3 = 0$, we get:-

localid="1645292125046" $\begin{array}{rcl}12\left(\frac{1}{9}\right)+5\sqrt{\frac{1}{9}}-3& =& \frac{4}{3}+\frac{5}{3}-3\\ \frac{4}{3}+\frac{5}{3}-3& =& \frac{9}{3}-3\\ \frac{9}{3}-3& =& 3-3\\ 3-3& =& 0\end{array}$

Since the value of $x=\frac{1}{9}$the equation satisfied the condition.

Hence, The solution of the equation is localid="1663940752402" $x=\frac{9}{16},x=\frac{1}{9}$.