Question

study paths tx math bridge - stage 2 unit test: radical expressions and quadratic equations radical expressions and quadratic equat unit: radical expressions and quadratic equations progress: the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer. rationalize the denominator and simplify: \\(\frac{3\sqrt{b}}{\sqrt{b}+2}\\) \\(\frac{3}{2}\\) \\(\frac{3b}{b + 2}\\) \\(\frac{9\sqrt{b}}{b - 4}\\) \\(\frac{3b - 6\sqrt{b}}{b - 4}\\)

Explanation:

Step1: Identify the conjugate of the denominator

The denominator is $\sqrt{b} + 2$, so its conjugate is $\sqrt{b} - 2$. We multiply the numerator and the denominator by this conjugate to rationalize the denominator.
$$\frac{3\sqrt{b}}{\sqrt{b}+2} \times \frac{\sqrt{b}-2}{\sqrt{b}-2}$$

Step2: Multiply the numerators and denominators

Multiply the numerators: $3\sqrt{b} \times (\sqrt{b} - 2) = 3\sqrt{b} \times \sqrt{b} - 3\sqrt{b} \times 2 = 3b - 6\sqrt{b}$
Multiply the denominators: $(\sqrt{b} + 2)(\sqrt{b} - 2)$ which is a difference of squares, so $(\sqrt{b})^2 - (2)^2 = b - 4$

Step3: Combine the results

Put the numerator and denominator together: $\frac{3b - 6\sqrt{b}}{b - 4}$

Answer:

$\frac{3b - 6\sqrt{b}}{b - 4}$ (the fourth option: $\boldsymbol{\frac{3b - 6\sqrt{b}}{b - 4}}$)