Rationalizing the Denominator

To rationalize the denominator means to eliminate any radical expressions in the denominator such as straight roots and cube roots. The key estimate is to multiply the original fraction by an appropriate respect, such that after reduction, the denominator no longer contains radicals. Note: In this lesson, we will only focus on square roots for the sake of simplicity. When the denominator is a monomial, the basic scheme is to apply the fact that

Conjugate of a Binomial

On the other handwriting, if the denominator is a binomial, the impression of conjugate solution comes in handy. The conjugate of a binomial is equal to the binomial itself, however, the middle sign is changed or switched. here are some examples of binomials with their represent conjugates. But more importantly, observe that the intersection of a given binomial and its conjugate is an saying without the group symbol . Let ’ s go over some examples !

Examples of How to Rationalize the Denominator

Example 1 : Rationalize the denominator \large { { 5 \over { \sqrt 2 } } }. Simplify further, if needed. The denominator contains a revolutionary expression, the squarely root of 2. Eliminate the radical at the bottomland by multiplying by itself which is \sqrt 2 since \sqrt 2 \cdot \sqrt 2 = \sqrt 4 = 2. however, by doing so we change the “ meaning ” or value of the master fraction. To balance it out, do the same thing on top by multiplying with the same value. What we are doing here is multiplying the master fraction by \large { { { \sqrt 2 } \over { \sqrt 2 } } } which is merely equivalent to 1. Remember that any number, when multiplied to 1, gives back the original number, frankincense, we change the imprint but not the original mean of the number itself ! so then the dim-witted solution to this problem is shown below . The final answer contains a denominator without a free radical symbol, and so we can claim that we have successfully rationalized it. Example 2 : Rationalize the denominator \large { { 6 \over { \sqrt 3 } } }. then simplify if necessary. observe that the denominator has a feather rout of 3. We have the need to rationalize it by getting rid of the revolutionary symbol . That means we should multiply the original fraction by \large { { { \sqrt 3 } \over { \sqrt 3 } } } . Yes, the root symbol at the buttocks is gone but we can still do something. Simplify foster by canceling out common factors . Example 3 : Rationalize \large { \sqrt { { { 27 } \over { 12 } } } }. What we have here is a square beginning of an entire fraction. The inaugural tone is to apply the Quotient Rule of Square Roots. This allows us to generate a fraction with a discrete numerator and denominator with radical symbols. QUOTIENT RULE OF SQUARE ROOTS By applying the rule above, we get a trouble that is more familiar to us . The “ new ” form of the given problem has a denominator of root 12, so we will multiply it by \large { { { \sqrt { 12 } } \over { \sqrt { 12 } } } } . As you can see, I did not multiply through the radicals on the numerator because the total will grow larger, frankincense more unmanageable to simplify. This way we should be able to simplify the numerator quite easily since the radicands are smaller when they are kept as is . Example 4 : simplify by rationalizing the denominator of \large { { { 7\sqrt { 10 } } \over { \sqrt 2 } } }. Multiply both the numerator and denominator by \sqrt 2. By doing so, the radical in the denominator should go away . We can ’ thyroxine stop here precisely so far because the revolutionary expression in the numerator contains a arrant hearty component. Let ’ s keep going with our reduction . cancel out common factors to get the final answer. You see that \large { { { 14 } \over 2 } = 7 }. Example 5 : simplify by rationalizing the denominator of \large { { { 6 – \sqrt 5 } \over { \sqrt 8 } } }. This meter the numerator contains a binomial. There ’ s in truth no difference in the approach because the denominator is however a monomial. We will multiply the original fraction by the denominator, \sqrt 8. Make sure to distribute \sqrt 8 into the terms inside the parenthesis.

The numerator can be far simplified because the radicands have arrant square factors. Observe that 8 = 4 \times 2 and 40 = 4 \times 10 where the factor 4\left ( { 4 = { 2^2 } } \right ) is a perfective feather number . Example 6 : Rationalize \large { { 2 \over { 3 + \sqrt 3 } } }. This trouble is a little bit different because the denominator is immediately a binomial, containing two terms. To get rid of the radical in the denominator, we are going to multiply the top and bottom by the conjugate of the given denominator. How do we get the conjugate of the denominator ? Take the same denominator but switch the middle sign . The middle gestural of the original denominator switches from plus to negative. Our option of multiplier that can rationalize the denominator is \large { { { 3 – \sqrt 3 } \over { 3 – \sqrt 3 } } }. here goes our solution . Example 7 : Rationalize \large { { 3 \over { 2 – \sqrt 2 } } }. The denominator has a negative sign in the center which makes the conjugate to have a convinced middle sign . The multiplier to use in order to rationalize the denominator is here goes our solution . No park factors between the numerator and denominator, therefore this is our final examination suffice. Example 8 : Rationalize \large { { { \sqrt 7 } \over { – 3 – \sqrt 7 } } }. The given denominator is
which makes its conjugate to be

• Given the problem:

• Multiply the numerator and denominator of the original fraction by the conjugate of the denominator.

• Distribute the numerators, and FOIL the denominators.

• The middle terms of the denominator will drop out because they are the “same” in values but opposite in signs.
• Simplify the roots of perfect square numbers, i.e.  \sqrt { 49 } = 7.

• Subtract the values in the denominator, 9 – 7 = 2.

• If possible, reduce the fraction to its lowest terms. It looks like there’s nothing to cancel out between the top and bottom. Therefore, keep this as our final answer.

Example 9 : Rationalize \large { { { \sqrt 6 – \sqrt 2 } \over { \sqrt 6 + \sqrt 2 } } }. Reverse the middle sign of the denominator to obtain its conjugate. That means, the conjugate of \sqrt 6 + \sqrt 2 is \sqrt 6 – \sqrt 2 .

• Multiply both top and bottom by the conjugate.

• Apply the FOIL method to expand the binomials.

• After expanding using FOIL, the middle terms of the numerator will be added, while the middle terms of the denominator will be canceled.

• Simplify the roots of perfect square numbers in every opportunity
• Add or subtract the whole numbers that come out after getting the square root of perfect square numbers.

• At this point, simplify

Remember to break it down as a merchandise wherein one of its factors is a perfect public square number .

• Obviously, 12 = 4 \times 3. This is a perfect choice of factors since the number 4 is a perfect square.

• Rewrite radical 12 as the product of the radicals of its factors.

• We know that the square root of 4 equals 2!

• Simplify by multiplication

• The numerator has a common factor of 4. That implies we could pull out a factor of 4 outside the parenthesis.

• Cancel common factors between the numerator and denominator

Example 10 : Rationalize \large { { { \sqrt 2 + \sqrt 8 } \over { – \sqrt 2 – \sqrt 8 } } } .

• Multiply the given fraction, both top and bottom, by the conjugate of the denominator.

• For now, multiply the binomials by placing them side by side.

• Expand the binomials using the FOIL method.
• Cancel out terms that are the “same” but opposite in signs.

• Get the exact values of the square roots of perfect square numbers.

• Perform the required arithmetic operations in both numerator and denominator.

• Cancel out common factors. The numerator and denominator is both divisible by 6

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