Math Problem Statement

not allowed to have square root in denominator

Solution

You're referring to a mathematical rule called rationalizing the denominator. In many cases, it's considered improper to leave a square root (or any radical) in the denominator of a fraction. The goal is to eliminate the square root from the denominator by multiplying both the numerator and the denominator by a suitable expression.

Example:

Suppose you have:

12\frac{1}{\sqrt{2}}

To rationalize the denominator, you multiply both the numerator and denominator by 2\sqrt{2} (the same square root in the denominator):

12×22=22\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}

Now the denominator is no longer a square root.

Another Example:

For a more complex expression, like:

35+2\frac{3}{\sqrt{5} + 2}

You would multiply both the numerator and denominator by the conjugate of the denominator, which in this case is 52\sqrt{5} - 2:

35+2×5252=3(52)(5)222=3(52)54=3(52)\frac{3}{\sqrt{5} + 2} \times \frac{\sqrt{5} - 2}{\sqrt{5} - 2} = \frac{3(\sqrt{5} - 2)}{(\sqrt{5})^2 - 2^2} = \frac{3(\sqrt{5} - 2)}{5 - 4} = 3(\sqrt{5} - 2)

So, the denominator is rationalized and the expression no longer has a square root in the denominator.


Would you like to go over any specific examples or details related to rationalizing the denominator?


Here are 5 related questions:

  1. What is the process of rationalizing a denominator with cube roots?
  2. Can you have a square root in the numerator?
  3. What happens if you rationalize a denominator with binomials that involve complex numbers?
  4. How do you rationalize denominators with higher-order roots (e.g., 4th roots)?
  5. How does rationalizing affect the overall value of an expression?

Tip: Always check whether you need to rationalize. Sometimes leaving a square root in the denominator is perfectly acceptable depending on the context, especially in advanced mathematics.

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Math Problem Analysis

Mathematical Concepts

Algebra
Rationalization of Denominators
Radicals

Formulas

\frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b}
\frac{a}{\sqrt{b} + c} \times \frac{\sqrt{b} - c}{\sqrt{b} - c} = \frac{a(\sqrt{b} - c)}{b - c^2}

Theorems

Rationalization of Denominators
Conjugate Pairs Theorem

Suitable Grade Level

Grades 8-10