Respuesta :
Answer:
[tex]\frac{4\sqrt{6x}}{2x}[/tex]
Explanation:
The problem we are given is
[tex]4\sqrt{\frac{3}{2x}}[/tex]
We can write the square root of a fraction as a fraction with a separate radical for the numerator and denominator; this gives us
[tex]4\times \frac{\sqrt{3}}{\sqrt{2x}}[/tex]
We can write the whole number 4 as the fraction 4/1; this gives us
[tex]\frac{4}{1}\times \frac{\sqrt{3}}{\sqrt{2x}}\\\\=\frac{4\sqrt{3}}{\sqrt{2x}}[/tex]
We now need to "rationalize the denominator." This means we need to cancel the square root in the denominator. In order to do this, we multiply both numerator and denominator by √(2x); this is because squaring a square root will cancel it:
[tex]\frac{4\sqrt{3}}{\sqrt{2x}}\times \frac{\sqrt{2x}}{\sqrt{2x}}\\\\=\frac{4\sqrt{3}\times \sqrt{2x}}{2x}[/tex]
When multiplying radicals, we can extend the radical over both factors:
[tex]\frac{4\sqrt{3} \times \sqrt{2x}}{2x}\\\\=\frac{4\sqrt{3\times 2x}}{2x}\\\\=\frac{4\sqrt{6x}}{2x}[/tex]
Answer:
The correct answer is ^4sqrt24x^3/2x or B on edge.
Step-by-step explanation:
