RATIONALIZING THE DENOMINATOR

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When a radical contains an expression that is not a perfect root, for example, the square root of 3 or cube root of 5, it is called an irrational number.

So, in order to rationalize the denominator, we have to get rid of all radicals that are in denominator.

The following steps are involved in rationalizing the denominator of rational expression.

Step 1 :

Multiply both numerator and denominator by a radical that will get rid of the radical in the denominator.

If the radical in the denominator is a square root, then we have to multiply by a square root that will give us a perfect square under the radical when multiplied by the denominator.

Step 2 :

Make sure all radicals are simplified.

Some radicals will already be in a simplified form, but we have to make sure that we simplify the ones that are not.

Step 3 :

Simplify the fraction if needed.

Be careful.  We cannot cancel out a factor that is on the outside of a radical with one that is on the inside of the radical.  In order to cancel out common factors, they have to be both inside the same radical or be both outside the radical.

Rationalize the denominator :

Example 1 :

ΒΉΒ²β„βˆšβ‚†

Solution :

ΒΉΒ²β„βˆšβ‚†

Multiply both numerator and denominator by βˆš6 to get rid of the radical in the denominator.

 = β½ΒΉΒ²Λ£βˆšβΆβΎβ„β‚βˆšβ‚†β‚“βˆšβ‚†β‚Ž

 = β½ΒΉΒ²Λ£βˆšβΆβΎβ„β‚†

= 2√6

Example 2 :

4√5/√10

Solution :

Simplify.

4√5/√10 = 4√5/√(2 β‹… 5)

= 4√5/(√2 β‹… βˆš5)

On the right side, cancel out βˆš5 in numerator and denominator.

= 4/√2

On the right side, multiply both numerator and denominator by βˆš2 to get rid of the radical in the denominator.

= (4 β‹… βˆš2)/(√2 β‹… βˆš2)

= 4√2/2

= 2√2

Example 3 :

5/√7

Solution :

Multiply both numerator and denominator by βˆš7 to get rid of the radical in the denominator.

5/√7 = (5 β‹… βˆš7)/(√7 β‹… βˆš7)

= 5√7 / 7

Example 4 :

12/√72

Solution :

Decompose 72 into prime factor using synthetic division.

√72 = βˆš(2 β‹… 2 β‹… 2 β‹… 3 β‹… 3)

= 2 β‹… 3 β‹… βˆš2

= 6√2

Then, we have

12/√72 = 12/6√2

Simplify.

= 2 / √2

On the right side, multiply both numerator and denominator by βˆš2 to get rid of the radical in the denominator.

= (2 β‹… βˆš2) β‹… (√2 β‹… βˆš2)

= 2√2/2

= √2

Example 5 :

1/(3 + √2)

Solution :

To get rid of the radical in denominator, multiply both numerator and denominator by the conjugate of (3 + βˆš2), that is by (3 - βˆš2).

1/(3 + √2) = [1 β‹… (3 - √2)]/[(3 + √2) β‹… (3 - √2)]

(3 - √2)/[(3 + √2) β‹… (3 - √2)]

Using the algebraic identity a2 - b2 = (a + b)(a - b), simplify the denominator on the right side.

= (3 - √2)/[32 - (√2)2]

= (3 - √2)/(9 - 2)

= (3 - √2)/7

Example 6 :

(1 - βˆš5)/(3 + √5)

Solution :

To get rid of the radical in denominator, multiply both numerator and denominator by the conjugate of (3 + βˆš5), that is by (3 - βˆš5).

(1 - βˆš5)/(3 + √5) = [(1 - βˆš5) β‹… (3 - √5)]/[(3 + √5) β‹… (3 - √5)]

Simplify.

= [3 βˆš5 - 3√5 + 5]/[32 - (√5)2]

= (8 - 4√5)/(9 - 5)

= 4(2 - √5)/4

= 2 - √5

Example 7 :

(√x + y)/(x - βˆšy)

Solution :

To get rid of the radical in denominator, multiply both numerator and denominator by the conjugate of (x - βˆšy), that is by (x + βˆšy).

(√x + y)/(x - √y) = [(√x + y) β‹… (x + √y)]/[(x - βˆšy) β‹… (x + √y)]

Simplify.

= [x√x + βˆšxy + xy + y√y]/[(x- (√y)2]

= [x√x + βˆšxy + xy + y√y]/(x- y2)

Example 8 :

3√(2/3a)

Solution :

3√(2/3a) = 3√2/3√3a

To rationalize the denominator in this case, multiply both numerator and denominator on the right side by the cube root of 9a2.

3√(2/3a) = [3√2 β‹… 3√(9a2)]/[3√3a β‹… 3√(9a2)]

Simplify.

3√(18a2)/3√(27a3)

3√(18a2)/3√(3 β‹… 3 β‹… 3 β‹… a β‹… a β‹… a)

3√(18a2)/3a

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