Simplify The Expression: $\[ \frac{-7}{\sqrt{5}+\sqrt{7}} \\]

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Introduction

Rationalizing the denominator is a crucial step in simplifying complex fractions, especially when dealing with square roots. In this article, we will delve into the process of simplifying the given expression: $\frac{-7}{\sqrt{5}+\sqrt{7}}$. We will explore the necessary steps to rationalize the denominator and simplify the expression.

Understanding the Problem

The given expression is a complex fraction with a square root in the denominator. To simplify this expression, we need to rationalize the denominator, which involves eliminating the square root from the denominator. This can be achieved by multiplying the numerator and denominator by the conjugate of the denominator.

Rationalizing the Denominator

To rationalize the denominator, we need to multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $\sqrt{5}+\sqrt{7}$ is $\sqrt{5}-\sqrt{7}$. By multiplying the numerator and denominator by the conjugate, we can eliminate the square root from the denominator.

Step 1: Multiply the Numerator and Denominator by the Conjugate

To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator:

$$\frac{-7}{\sqrt{5}+\sqrt{7}} \cdot \frac{\sqrt{5}-\sqrt{7}}{\sqrt{5}-\sqrt{7}}$$

Step 2: Simplify the Expression

Now, we simplify the expression by multiplying the numerator and denominator:

$$\frac{-7(\sqrt{5}-\sqrt{7})}{(\sqrt{5}+\sqrt{7})(\sqrt{5}-\sqrt{7})}$$

Step 3: Apply the Difference of Squares Formula

To simplify the denominator, we apply the difference of squares formula:

$$(a+b)(a-b) = a^2 - b^2$$

In this case, the denominator becomes:

$$(\sqrt{5}+\sqrt{7})(\sqrt{5}-\sqrt{7}) = \sqrt{5}^2 - \sqrt{7}^2 = 5 - 7 = -2$$

Step 4: Simplify the Expression

Now, we simplify the expression by substituting the simplified denominator:

$$\frac{-7(\sqrt{5}-\sqrt{7})}{-2}$$

Step 5: Distribute the Negative Sign

To simplify the expression, we distribute the negative sign to the terms inside the parentheses:

$$\frac{7(\sqrt{7}-\sqrt{5})}{2}$$

Conclusion

In this article, we simplified the given expression: $\frac{-7}{\sqrt{5}+\sqrt{7}}$. We rationalized the denominator by multiplying the numerator and denominator by the conjugate of the denominator. By applying the difference of squares formula and simplifying the expression, we arrived at the final simplified expression: $\frac{7(\sqrt{7}-\sqrt{5})}{2}$.

Final Answer

The final answer is: $\boxed{\frac{7(\sqrt{7}-\sqrt{5})}{2}}$

Related Topics

• Rationalizing the denominator
• Simplifying complex fractions
• Square roots
• Conjugate of a binomial

Example Problems

• Simplify the expression: $\frac{3}{\sqrt{2}+\sqrt{3}}$
• Rationalize the denominator: $\frac{2}{\sqrt{3}-\sqrt{2}}$
• Simplify the expression: $\frac{5}{\sqrt{7}-\sqrt{4}}$

Practice Problems

• Simplify the expression: $\frac{4}{\sqrt{9}+\sqrt{6}}$
• Rationalize the denominator: $\frac{3}{\sqrt{8}-\sqrt{5}}$
• Simplify the expression: $\frac{2}{\sqrt{3}+\sqrt{9}}$

Solutions to Practice Problems

• Simplify the expression: $\frac{4}{\sqrt{9}+\sqrt{6}} = \frac{4(\sqrt{9}-\sqrt{6})}{(\sqrt{9}+\sqrt{6})(\sqrt{9}-\sqrt{6})} = \frac{4(\sqrt{9}-\sqrt{6})}{9-6} = \frac{4(\sqrt{9}-\sqrt{6})}{3} = \frac{4\sqrt{9}-4\sqrt{6}}{3} = \frac{12-4\sqrt{6}}{3} = \frac{4(3-\sqrt{6})}{3} = \frac{4}{3}(3-\sqrt{6})$
• Rationalize the denominator: $\frac{3}{\sqrt{8}-\sqrt{5}} = \frac{3}{\sqrt{8}-\sqrt{5}} \cdot \frac{\sqrt{8}+\sqrt{5}}{\sqrt{8}+\sqrt{5}} = \frac{3(\sqrt{8}+\sqrt{5})}{(\sqrt{8}-\sqrt{5})(\sqrt{8}+\sqrt{5})} = \frac{3(\sqrt{8}+\sqrt{5})}{8-5} = \frac{3(\sqrt{8}+\sqrt{5})}{3} = \sqrt{8}+\sqrt{5}$
• Simplify the expression: $\frac{2}{\sqrt{3}+\sqrt{9}} = \frac{2}{\sqrt{3}+\sqrt{9}} \cdot \frac{\sqrt{3}-\sqrt{9}}{\sqrt{3}-\sqrt{9}} = \frac{2(\sqrt{3}-\sqrt{9})}{(\sqrt{3}+\sqrt{9})(\sqrt{3}-\sqrt{9})} = \frac{2(\sqrt{3}-\sqrt{9})}{3-9} = \frac{2(\sqrt{3}-\sqrt{9})}{-6} = -\frac{1}{3}(\sqrt{3}-\sqrt{9}) = -\frac{1}{3}(\sqrt{3}-3)$
  

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Introduction

In our previous article, we simplified the expression: $\frac{-7}{\sqrt{5}+\sqrt{7}}$. We rationalized the denominator by multiplying the numerator and denominator by the conjugate of the denominator. In this article, we will answer some frequently asked questions related to the simplification of the expression.

Q&A

Q: What is the conjugate of a binomial?

A: The conjugate of a binomial is obtained by changing the sign of the second term. For example, the conjugate of $a+b$ is $a-b$, and the conjugate of $a-b$ is $a+b$.

Q: Why do we need to rationalize the denominator?

A: We need to rationalize the denominator to eliminate the square root from the denominator. This is necessary because square roots cannot be simplified further, and we need to simplify the expression to its simplest form.

Q: How do we rationalize the denominator?

A: To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator. This eliminates the square root from the denominator and simplifies the expression.

Q: What is the difference of squares formula?

A: The difference of squares formula is: $(a+b)(a-b) = a^2 - b^2$. This formula is used to simplify the denominator when rationalizing the denominator.

Q: Can we simplify the expression further?

A: Yes, we can simplify the expression further by distributing the negative sign to the terms inside the parentheses.

Q: What is the final simplified expression?

A: The final simplified expression is: $\frac{7(\sqrt{7}-\sqrt{5})}{2}$.

Example Problems with Solutions

• Simplify the expression: $\frac{3}{\sqrt{2}+\sqrt{3}}$
  • Solution: $\frac{3}{\sqrt{2}+\sqrt{3}} \cdot \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}} = \frac{3(\sqrt{2}-\sqrt{3})}{2-3} = -3(\sqrt{2}-\sqrt{3})$
• Rationalize the denominator: $\frac{2}{\sqrt{3}-\sqrt{2}}$
  • Solution: $\frac{2}{\sqrt{3}-\sqrt{2}} \cdot \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} = \frac{2(\sqrt{3}+\sqrt{2})}{3-2} = 2(\sqrt{3}+\sqrt{2})$
• Simplify the expression: $\frac{5}{\sqrt{7}-\sqrt{4}}$
  • Solution: $\frac{5}{\sqrt{7}-\sqrt{4}} \cdot \frac{\sqrt{7}+\sqrt{4}}{\sqrt{7}+\sqrt{4}} = \frac{5(\sqrt{7}+\sqrt{4})}{7-4} = \frac{5(\sqrt{7}+2)}{3}$

Practice Problems

• Simplify the expression: $\frac{4}{\sqrt{9}+\sqrt{6}}$
• Rationalize the denominator: $\frac{3}{\sqrt{8}-\sqrt{5}}$
• Simplify the expression: $\frac{2}{\sqrt{3}+\sqrt{9}}$

Solutions to Practice Problems

• Simplify the expression: $\frac{4}{\sqrt{9}+\sqrt{6}} = \frac{4(\sqrt{9}-\sqrt{6})}{(\sqrt{9}+\sqrt{6})(\sqrt{9}-\sqrt{6})} = \frac{4(\sqrt{9}-\sqrt{6})}{9-6} = \frac{4(\sqrt{9}-\sqrt{6})}{3} = \frac{4\sqrt{9}-4\sqrt{6}}{3} = \frac{12-4\sqrt{6}}{3} = \frac{4(3-\sqrt{6})}{3} = \frac{4}{3}(3-\sqrt{6})$
• Rationalize the denominator: $\frac{3}{\sqrt{8}-\sqrt{5}} = \frac{3}{\sqrt{8}-\sqrt{5}} \cdot \frac{\sqrt{8}+\sqrt{5}}{\sqrt{8}+\sqrt{5}} = \frac{3(\sqrt{8}+\sqrt{5})}{(\sqrt{8}-\sqrt{5})(\sqrt{8}+\sqrt{5})} = \frac{3(\sqrt{8}+\sqrt{5})}{8-5} = \frac{3(\sqrt{8}+\sqrt{5})}{3} = \sqrt{8}+\sqrt{5}$
• Simplify the expression: $\frac{2}{\sqrt{3}+\sqrt{9}} = \frac{2}{\sqrt{3}+\sqrt{9}} \cdot \frac{\sqrt{3}-\sqrt{9}}{\sqrt{3}-\sqrt{9}} = \frac{2(\sqrt{3}-\sqrt{9})}{(\sqrt{3}+\sqrt{9})(\sqrt{3}-\sqrt{9})} = \frac{2(\sqrt{3}-\sqrt{9})}{3-9} = \frac{2(\sqrt{3}-\sqrt{9})}{-6} = -\frac{1}{3}(\sqrt{3}-\sqrt{9}) = -\frac{1}{3}(\sqrt{3}-3)$