Simplifying Complex Fractions with Square Roots

A fraction has two parts: numerator & denominator; the numerator is the number above the line, and the denominator is the number below the line. In a fraction, the line or slash that divides the numerator and denominator denotes division.

Complex Fractions

A fraction with fractions in the denominator and numerator, or both, is called a complicated fraction. A complex rational expression is a complex fraction that contains a variable.

Other examples include: \frac{3}{\frac{1}{2}}, \frac{69}{\sqrt{420}}, etc.

Steps to simplifying complex fractions

• Multiply both the numerator and denominator by the radical that can remove the radical in the denominator. For example, in the expression \frac{2}{\sqrt3} , the common radical to be multiplied in order to rationalize the denominator is √3.
• Evaluate the expression, if possible, by multiplying the terms and further simplifying the numerator and denominator by taking out any common factors.

Example: Rationalize the denominator: \frac{4}{\sqrt6}.

Solution: 

Given: \frac{4}{\sqrt6}   

Clearly, in order to rationalize the denominator, √6 needs to be multiplied with both the numerator and numerator.

= \frac{4}{\sqrt6}×\frac{\sqrt{6}}{\sqrt6}

= \frac{4\sqrt{6}}{(\sqrt6)^2}

= \frac{4\sqrt{6}}{6}

= 2√6/3

Practice Problems on Simplifying Complex Fractions with Square Roots

Question 1: Rationalize the denominator: \frac{4\sqrt3}{2+\sqrt5}.

Solution:

Multiply both the numerator and denominator with 2 − √5 to rationalize the denominator.

\frac{4\sqrt3}{2+\sqrt5}×\frac{2-\sqrt5}{2-\sqrt5}

Using the identity (a + b)(a − b) = a² − b², we have:

= \frac{4\sqrt3(2-\sqrt5)}{2^2-(\sqrt5)^2}\\=\frac{8\sqrt3-4\sqrt{15}}{4-5}\\=\frac{8\sqrt3-4\sqrt{15}}{-1}\\=4\sqrt{15}-8\sqrt{3}

Question 2: Rationalize the denominator: \frac{6+\sqrt3}{4-\sqrt3}.

Solution:

Multiply both the numerator and denominator with 4 + √3 to rationalize the denominator.

\frac{6+\sqrt3}{4-\sqrt3} ×\frac{4+\sqrt3}{4+\sqrt3}

Using the identity (a + b)(a − b) = a² − b², we have:

= \frac{6+\sqrt3(4+\sqrt3)}{4^2-(\sqrt3)^2}\\=\frac{24+10\sqrt3+3}{16-3}\\=\frac{27+10\sqrt3}{13}

Question 3: Rationalize the denominator: \frac{1}{\sqrt5-2}.

Solution:

To rationalise the denominator of \frac{1}{\sqrt{5}-2}

Multiply numerator and denominator by the conjugtae of the denominator, \sqrt{5}+2

\frac{1}{\sqrt{5}-2} \times \frac{\sqrt{5}+2}{\sqrt{5}+2}

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

Now simplify the denominator:

(√5​)² - 2² = 5 - 4 = 1

So the expression becomes:

\frac{\sqrt{5}+2}{1}

= \sqrt{5}+2

Question 4: Rationalize the denominator: \frac{6}{3+\sqrt7}.

Solution:

Multiply both the numerator and denominator with 3 - √7 to rationalize the denominator.

\frac{6}{3+\sqrt7}×\frac{3-\sqrt7}{3-\sqrt7}

Using the identity (a + b)(a − b) = a² − b², we have:

= \frac{6(3-\sqrt7)}{3^2-(\sqrt7)^2}\\=\frac{18-6\sqrt7}{9-7}\\=\frac{18-6\sqrt7}{2}\\=9-3\sqrt7

Question 5: Rationalize the denominator: \frac{2}{\sqrt5-2}.

Solution:

Multiply both the numerator and denominator with √5 + 2 to rationalize the denominator.

\frac{2}{2+\sqrt5}×\frac{√5 + 2}{√5 + 2}

Using the identity (a + b)(a − b) = a² − b², we have:

= \frac{2\sqrt5+4}{(\sqrt5)^2-2^2}

Simplify the denominator: 5 - 4 = 1

\frac{2\sqrt5+4}{1}

Question 6: Rationalize the denominator: \frac{7}{\sqrt5-2}.

Solution:

Multiply both the numerator and denominator with √5 + 2 to rationalize the denominator.

\frac{7}{2+\sqrt5}×\frac{√5 + 2}{√5 + 2}

Using the identity (a + b)(a − b) = a² − b², we have:

= \frac{7\sqrt5+14}{(\sqrt5)^2-2^2}

=\frac{2\sqrt5+4}{5-4}

=2\sqrt{5}+4

Question 7: Rationalize the denominator: \frac{3}{\sqrt5}.

Solution:

Given: \frac{3}{\sqrt5}

Clearly, in order to rationalize the denominator, √6 needs to be multiplied with both the numerator and numerator.

= \frac{3}{\sqrt5}×\frac{\sqrt{5}}{\sqrt5}\\=\frac{3\sqrt{5}}{(\sqrt5)^2}\\=\frac{3\sqrt{5}}{5}\

Related Articles:

• Simple and Complex Fractions
• How to Simply Complex Fractions
• How to Add Complex Fractions