Rationalize the denominator.
$\frac{6+\sqrt{x}}{-\sqrt{x}}$

• A. $\frac{6\sqrt{x}}{-x}-1$
• B. $6\sqrt{x}+x$
• C. $\frac{6\sqrt{x}}{x}+1$
• D. $\frac{7\sqrt{x}}{-x}$

Answer: (A)

Verified step by step guidance

1. Identify the given expression: \(\frac{6 + \sqrt{x}}{-\sqrt{x}}\). Our goal is to rationalize the denominator, which means eliminating the square root from the denominator.
2. Multiply both the numerator and the denominator by the conjugate of the denominator. Since the denominator is \(-\sqrt{x}\), its conjugate is simply \(-\sqrt{x}\) itself (because it is a single term). So multiply numerator and denominator by \(-\sqrt{x}\):
3. \[\frac{6 + \sqrt{x}}{-\sqrt{x}} \times \frac{-\sqrt{x}}{-\sqrt{x}}\]
4. Apply the multiplication in numerator and denominator separately:
5. - Numerator: \((6 + \sqrt{x})(-\sqrt{x}) = 6 \times (-\sqrt{x}) + \sqrt{x} \times (-\sqrt{x})\)
6. - Denominator: \((-\sqrt{x}) \times (-\sqrt{x}) = (-1)^2 \times (\sqrt{x})^2 = x\)
7. Simplify the numerator terms:
8. - \$6 \times (-\sqrt{x}) = -6\sqrt{x}$
9. - \(\sqrt{x} \times (-\sqrt{x}) = -x\)
10. So the numerator becomes \(-6\sqrt{x} - x\) and the denominator is \(x\). Write the fraction as:
11. \[\frac{-6\sqrt{x} - x}{x}\]
12. Finally, you can separate the terms in the numerator over the denominator:
13. \[\frac{-6\sqrt{x}}{x} - \frac{x}{x}\]
14. Simplify \(\frac{x}{x}\) to 1, so the expression becomes:
15. \[-\frac{6\sqrt{x}}{x} - 1\]
16. This is the rationalized form of the original expression.