First, let's make the following substitutions:
[tex]\begin{gathered} y=\sqrt{x} \\ y^2=x \end{gathered}[/tex]
Then, substituting in the given equation, we have:
[tex]2y^2-9y+4=0[/tex]
Now, let's solve using the quadratic formula:
[tex]\begin{gathered} a=2,b=-9,c=4\\ \\ y=\frac{-b±\sqrt{b^2-4ac}}{2a}\\ \\ y=\frac{9±\sqrt{81-32}}{4}\\ \\ y_1=\frac{9+7}{4}=4\\ \\ y_2=\frac{9-7}{4}=\frac{1}{2} \end{gathered}[/tex]
Now, let's calculate the value of x:
[tex]\begin{gathered} y=\sqrt{x}\\ \\ \begin{cases}4=\sqrt{x}\rightarrow x=16 \\ \frac{1}{2}={\sqrt{x}}\rightarrow x=\frac{1}{4}\end{cases} \end{gathered}[/tex]
Therefore the solutions are x = 1/4 and x = 16.
