Simplify 3/( square root of x+ square root of 5)
The question is asking for the simplification of the mathematical expression 3 divided by the sum of the square root of 'x' and the square root of 5. This involves algebraic manipulation to eliminate the square root from the denominator of the fraction, typically through a process called "rationalizing the denominator," where both the numerator and denominator of the fraction are multiplied by a conjugate of the denominator to create a rational number in the denominator.
$\frac{3}{\sqrt{x} + \sqrt{5}}$
Answer
Solution:
Step 1:
Rationalize the denominator by multiplying the fraction $\frac{3}{\sqrt{x} + \sqrt{5}}$ by the conjugate $\frac{\sqrt{x} - \sqrt{5}}{\sqrt{x} - \sqrt{5}}$ to get $\frac{3}{\sqrt{x} + \sqrt{5}} \cdot \frac{\sqrt{x} - \sqrt{5}}{\sqrt{x} - \sqrt{5}}$.
Step 2:
Combine the numerator terms by distributing the 3 across $\sqrt{x} - \sqrt{5}$, resulting in $\frac{3(\sqrt{x} - \sqrt{5})}{(\sqrt{x} + \sqrt{5})(\sqrt{x} - \sqrt{5})}$.
Step 3:
Apply the difference of squares to the denominator, which simplifies to $\sqrt{x}^2 - (\sqrt{5})^2$.
Step 4:
Complete the simplification to obtain $\frac{3(\sqrt{x} - \sqrt{5})}{x - 5}$.
Knowledge Notes:
To simplify a fraction where the denominator contains a radical, we often multiply the fraction by a form of 1 that will eliminate the radical in the denominator. This process is known as rationalizing the denominator.
The conjugate of a binomial expression $\sqrt{a} + \sqrt{b}$ is $\sqrt{a} - \sqrt{b}$. When we multiply a binomial by its conjugate, the result is the difference of squares, which is a special product that eliminates the radical: $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$.
The FOIL method stands for First, Outer, Inner, Last, referring to the order in which we multiply terms when expanding the product of two binomials. However, when multiplying conjugates, the middle terms cancel out, leaving only the first and last terms.
In the given problem, after rationalizing the denominator and applying the difference of squares, we get a simplified expression where the radicals are eliminated from the denominator, making it easier to handle and understand.
