Simplify The Expression:$ \frac{9}{\sqrt{3}+\sqrt{7}} $

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Introduction

In mathematics, simplifying complex expressions is an essential skill that helps us solve problems efficiently and accurately. One such expression is 93+7\frac{9}{\sqrt{3}+\sqrt{7}}. In this article, we will explore the steps involved in simplifying this expression and provide a clear understanding of the process.

Understanding the Expression

The given expression is 93+7\frac{9}{\sqrt{3}+\sqrt{7}}. To simplify this expression, we need to rationalize the denominator, which means removing the square roots from the denominator.

Rationalizing the Denominator

To rationalize the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of 3+7\sqrt{3}+\sqrt{7} is 3βˆ’7\sqrt{3}-\sqrt{7}.

\frac{9}{\sqrt{3}+\sqrt{7}} \cdot \frac{\sqrt{3}-\sqrt{7}}{\sqrt{3}-\sqrt{7}}

Expanding the Expression

Now, we expand the expression by multiplying the numerator and the denominator.

\frac{9(\sqrt{3}-\sqrt{7})}{(\sqrt{3}+\sqrt{7})(\sqrt{3}-\sqrt{7})}

Simplifying the Expression

To simplify the expression, we use the difference of squares formula: (a+b)(aβˆ’b)=a2βˆ’b2(a+b)(a-b) = a^2 - b^2.

\frac{9(\sqrt{3}-\sqrt{7})}{(\sqrt{3})^2 - (\sqrt{7})^2}

Further Simplification

Now, we simplify the expression further by evaluating the squares.

\frac{9(\sqrt{3}-\sqrt{7})}{3 - 7}

Final Simplification

Finally, we simplify the expression by evaluating the subtraction.

\frac{9(\sqrt{3}-\sqrt{7})}{-4}

Multiplying the Numerator

To simplify the expression further, we multiply the numerator by βˆ’1-1 to get rid of the negative sign in the denominator.

\frac{-9(\sqrt{3}-\sqrt{7})}{4}

Simplifying the Expression

Now, we simplify the expression by distributing the negative sign to the terms inside the parentheses.

\frac{-9\sqrt{3}+9\sqrt{7}}{4}

Conclusion

In this article, we simplified the complex expression 93+7\frac{9}{\sqrt{3}+\sqrt{7}} by rationalizing the denominator and using the difference of squares formula. We also multiplied the numerator by βˆ’1-1 to get rid of the negative sign in the denominator. The final simplified expression is βˆ’93+974\frac{-9\sqrt{3}+9\sqrt{7}}{4}.

Key Takeaways

  • To simplify a complex expression, we need to rationalize the denominator.
  • The conjugate of a binomial expression is obtained by changing the sign of the second term.
  • The difference of squares formula is (a+b)(aβˆ’b)=a2βˆ’b2(a+b)(a-b) = a^2 - b^2.
  • To simplify an expression, we need to evaluate the squares and the subtraction.

Practice Problems

  1. Simplify the expression 42+5\frac{4}{\sqrt{2}+\sqrt{5}}.
  2. Simplify the expression 92βˆ’3\frac{9}{\sqrt{2}-\sqrt{3}}.
  3. Simplify the expression 163+4\frac{16}{\sqrt{3}+\sqrt{4}}.

Answer Key

  1. 4(2βˆ’5)(2)2βˆ’(5)2=4(2βˆ’5)2βˆ’5=4(2βˆ’5)βˆ’3=βˆ’4(2βˆ’5)3=βˆ’42+453\frac{4(\sqrt{2}-\sqrt{5})}{(\sqrt{2})^2 - (\sqrt{5})^2} = \frac{4(\sqrt{2}-\sqrt{5})}{2 - 5} = \frac{4(\sqrt{2}-\sqrt{5})}{-3} = \frac{-4(\sqrt{2}-\sqrt{5})}{3} = \frac{-4\sqrt{2}+4\sqrt{5}}{3}
  2. 9(2+3)(2)2βˆ’(3)2=9(2+3)2βˆ’3=9(2+3)βˆ’1=βˆ’9(2+3)=βˆ’92βˆ’93\frac{9(\sqrt{2}+\sqrt{3})}{(\sqrt{2})^2 - (\sqrt{3})^2} = \frac{9(\sqrt{2}+\sqrt{3})}{2 - 3} = \frac{9(\sqrt{2}+\sqrt{3})}{-1} = -9(\sqrt{2}+\sqrt{3}) = -9\sqrt{2}-9\sqrt{3}
  3. 163+4=163+2=16(3βˆ’2)(3)2βˆ’22=16(3βˆ’2)3βˆ’4=16(3βˆ’2)βˆ’1=βˆ’16(3βˆ’2)=16(2βˆ’3)=32βˆ’163\frac{16}{\sqrt{3}+\sqrt{4}} = \frac{16}{\sqrt{3}+2} = \frac{16(\sqrt{3}-2)}{(\sqrt{3})^2 - 2^2} = \frac{16(\sqrt{3}-2)}{3 - 4} = \frac{16(\sqrt{3}-2)}{-1} = -16(\sqrt{3}-2) = 16(2-\sqrt{3}) = 32-16\sqrt{3}
    Simplifying Complex Expressions: A Q&A Guide =====================================================

Introduction

In our previous article, we explored the steps involved in simplifying complex expressions, including rationalizing the denominator and using the difference of squares formula. In this article, we will answer some frequently asked questions (FAQs) related to simplifying complex expressions.

Q: What is rationalizing the denominator?

A: Rationalizing the denominator is the process of removing the square roots from the denominator of a fraction. This is done by multiplying both the numerator and the denominator by the conjugate of the denominator.

Q: What is the conjugate of a binomial expression?

A: The conjugate of a binomial expression is obtained by changing the sign of the second term. For example, the conjugate of 3+7\sqrt{3}+\sqrt{7} is 3βˆ’7\sqrt{3}-\sqrt{7}.

Q: How do I simplify a complex expression?

A: To simplify a complex expression, you need to follow these steps:

  1. Rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator.
  2. Expand the expression by multiplying the numerator and the denominator.
  3. Simplify the expression by evaluating the squares and the subtraction.
  4. Finally, simplify the expression by distributing the negative sign to the terms inside the parentheses.

Q: What is the difference of squares formula?

A: The difference of squares formula is (a+b)(aβˆ’b)=a2βˆ’b2(a+b)(a-b) = a^2 - b^2. This formula is used to simplify expressions that involve the product of two binomials.

Q: How do I use the difference of squares formula?

A: To use the difference of squares formula, you need to identify the two binomials in the expression and multiply them together. Then, you need to simplify the resulting expression by evaluating the squares and the subtraction.

Q: What are some common mistakes to avoid when simplifying complex expressions?

A: Some common mistakes to avoid when simplifying complex expressions include:

  • Not rationalizing the denominator
  • Not using the difference of squares formula
  • Not evaluating the squares and the subtraction
  • Not distributing the negative sign to the terms inside the parentheses

Q: How do I practice simplifying complex expressions?

A: To practice simplifying complex expressions, you can try the following:

  • Start with simple expressions and gradually move on to more complex ones.
  • Use online resources, such as math websites and apps, to practice simplifying complex expressions.
  • Work with a tutor or a teacher to get personalized feedback and guidance.
  • Join a study group or a math club to practice simplifying complex expressions with others.

Q: What are some real-world applications of simplifying complex expressions?

A: Simplifying complex expressions has many real-world applications, including:

  • Physics: Simplifying complex expressions is essential in physics, where you need to solve equations that involve complex numbers and variables.
  • Engineering: Simplifying complex expressions is also essential in engineering, where you need to design and analyze complex systems and structures.
  • Computer Science: Simplifying complex expressions is also essential in computer science, where you need to write algorithms and programs that involve complex mathematical operations.

Conclusion

In this article, we answered some frequently asked questions (FAQs) related to simplifying complex expressions. We also provided some tips and resources for practicing simplifying complex expressions. By following these tips and resources, you can become proficient in simplifying complex expressions and apply them to real-world problems.

Key Takeaways

  • Rationalizing the denominator is the process of removing the square roots from the denominator of a fraction.
  • The conjugate of a binomial expression is obtained by changing the sign of the second term.
  • The difference of squares formula is (a+b)(aβˆ’b)=a2βˆ’b2(a+b)(a-b) = a^2 - b^2.
  • To simplify a complex expression, you need to follow the steps outlined in this article.

Practice Problems

  1. Simplify the expression 42+5\frac{4}{\sqrt{2}+\sqrt{5}}.
  2. Simplify the expression 92βˆ’3\frac{9}{\sqrt{2}-\sqrt{3}}.
  3. Simplify the expression 163+4\frac{16}{\sqrt{3}+\sqrt{4}}.

Answer Key

  1. 4(2βˆ’5)(2)2βˆ’(5)2=4(2βˆ’5)2βˆ’5=4(2βˆ’5)βˆ’3=βˆ’4(2βˆ’5)3=βˆ’42+453\frac{4(\sqrt{2}-\sqrt{5})}{(\sqrt{2})^2 - (\sqrt{5})^2} = \frac{4(\sqrt{2}-\sqrt{5})}{2 - 5} = \frac{4(\sqrt{2}-\sqrt{5})}{-3} = \frac{-4(\sqrt{2}-\sqrt{5})}{3} = \frac{-4\sqrt{2}+4\sqrt{5}}{3}
  2. 9(2+3)(2)2βˆ’(3)2=9(2+3)2βˆ’3=9(2+3)βˆ’1=βˆ’9(2+3)=βˆ’92βˆ’93\frac{9(\sqrt{2}+\sqrt{3})}{(\sqrt{2})^2 - (\sqrt{3})^2} = \frac{9(\sqrt{2}+\sqrt{3})}{2 - 3} = \frac{9(\sqrt{2}+\sqrt{3})}{-1} = -9(\sqrt{2}+\sqrt{3}) = -9\sqrt{2}-9\sqrt{3}
  3. 163+4=163+2=16(3βˆ’2)(3)2βˆ’22=16(3βˆ’2)3βˆ’4=16(3βˆ’2)βˆ’1=βˆ’16(3βˆ’2)=16(2βˆ’3)=32βˆ’163\frac{16}{\sqrt{3}+\sqrt{4}} = \frac{16}{\sqrt{3}+2} = \frac{16(\sqrt{3}-2)}{(\sqrt{3})^2 - 2^2} = \frac{16(\sqrt{3}-2)}{3 - 4} = \frac{16(\sqrt{3}-2)}{-1} = -16(\sqrt{3}-2) = 16(2-\sqrt{3}) = 32-16\sqrt{3}