The given equation is
[tex]\sqrt[]{3\text{ - 2x}}\text{ - 4x = 0}[/tex]The first step is to add 4x to both sides of the equation. We have
[tex]\begin{gathered} \sqrt[]{3\text{ - 2x}}\text{ - 4x + 4x = 0 + 4x} \\ \sqrt[]{3\text{ - 2x}}\text{ = 4x} \\ \text{Squaring both sides of the equation, we have} \\ (\sqrt[]{3-2x)}^2=(4x)^2 \\ 3-2x=16x^2 \end{gathered}[/tex]3 - 2x = 16x^2
Adding 2x to both sides of the equation, we have
3 - 2x + 2x = 16x^2 + 2x
3 = 16x^2 + 2x
Subtracting 3 from both sides of the equation, we have
3 - 3 = 16x^2 + 2x - 3
0 = 16x^2 + 2x - 3
16x^2 + 2x - 3 = 0
This is a quadratic equation. We would solve for x by applying the method of factorisation. The first step is to multiply the first and last terms. We have 16x^2 * - 3 = - 48x^2. We would find two terms such that their sum or difference is 2x and their product is - 48x^2. The terms are 8x and - 6x. By replacing 2x with with 8x - 6x in the equation, we have
16x^2 + 8x - 6x - 3 = 0
By factorising, we have
8x(2x + 1) - 3(2x + 1) = 0
Since 2x + 1 is common, we have
(2x + 1)(8x - 3) = 0
2x + 1 = 0 or 8x - 3 = 0
2x = - 1 or 8x = 3
x = - 1/2 or x = 3/8
We would substitute these values in the original equation to check. We have
[tex]\begin{gathered} For\text{ x = }-\text{ }\frac{1}{2} \\ \sqrt[]{3\text{ - 2}\times-\frac{1}{2}}\text{ - 4}\times-\text{ }\frac{1}{2}\text{ = 0} \\ \sqrt[]{3\text{ - - 1}}\text{ + 2 = 0} \\ \sqrt[]{4}\text{ + 2 = 0} \\ 2\text{ + 2 }\ne0 \end{gathered}[/tex][tex]\begin{gathered} \text{For x = }\frac{3}{8} \\ \sqrt[]{3\text{ - 2}\times\frac{3}{8}}\text{ - 4}\times\frac{3}{8}\text{ = 0} \\ \sqrt[]{3\text{ - }\frac{3}{4}}\text{ - }\frac{3}{2}=\text{ 0} \\ \sqrt[]{\frac{9}{4}}\text{ - }\frac{3}{2}\text{ = 0} \\ \frac{3}{2}\text{ - }\frac{3}{2}\text{ = 0} \end{gathered}[/tex]The solution is x = 3/8
