Question

By multiplying both numerator and denominator of \(\frac{1}{a+b \sqrt{c}}\) by \(a-b \sqrt{c}\) show that $$ \frac{1}{a+b \sqrt{c}}=\frac{a-b \sqrt{c}}{a^{2}-b^{2} c} $$ Use this approach to show that $$ \frac{1}{2+\sqrt{3}}=2-\sqrt{3} $$

Short answer

Answer: The simplified form of the fraction is $\frac{a-b\sqrt{c}}{a^2-b^2c}$. When applied to the example, we get $\frac{1}{2+\sqrt{3}} = 2-\sqrt{3}$.

Step-by-step solution

Multiply the numerator and the denominator by \((a-b\sqrt{c})\)

We multiply both the numerator and the denominator of \(\frac{1}{a+b\sqrt{c}}\) by \(a-b\sqrt{c}\): $$ \frac{1}{a+b\sqrt{c}}\times\frac{a-b\sqrt{c}}{a-b\sqrt{c}} $$

Simplify the result

We perform the multiplication and simplify the fraction obtained in Step 1: $$ \frac{1(a-b\sqrt{c})}{(a+b\sqrt{c})(a-b\sqrt{c})} $$ $$ \frac{a-b\sqrt{c}}{a^2-(b\sqrt{c})^2} $$ By squaring \(b\sqrt{c}\), we obtain: $$ \frac{a-b\sqrt{c}}{a^2-b^2c} $$

Apply the approach to the specific case

We apply the approach to the specific example of \(\frac{1}{2+\sqrt{3}}\). In this case, we have \(a = 2\), \(b = 1\), and \(c = 3\). We can then apply the result obtained in Step 2: $$ \frac{2-1\sqrt{3}}{(2)^2-(1)^2(3)} $$ $$ \frac{2-\sqrt{3}}{4-3} $$ $$ 2-\sqrt{3} $$ Thus, we have shown that $$ \frac{1}{2+\sqrt{3}} = 2-\sqrt{3}. $$

Key concepts

Algebraic Fractions

Algebraic fractions are expressions that have a polynomial in both the numerator and the denominator. They are quite similar to numerical fractions but instead of numbers, we deal with algebraic expressions. A typical example could be \( \frac{x+2}{x-3} \). Just like numerical fractions, algebraic fractions can be simplified by factoring and reducing common terms.

Algebraic fractions become essential when dealing with expressions involving variables. They can be added, subtracted, multiplied, and divided using similar principles as ordinary fractions. However, they often involve more complex manipulations.

• To simplify, find the greatest common factor in both numerator and denominator.
• Factor the expressions if possible.
• Cancel out common factors.

Handling these expressions involves ensuring the denominator is not equal to zero, as division by zero is undefined. A good grasp of algebraic fractions is key for more advanced topics like rationalizing denominators.

Surds

Surds refer to irrational expressions containing a square root, cube root, or other roots that cannot be simplified into rational numbers. Examples include \( \sqrt{2} \) and \( \sqrt{3} \). Working with surds is common in algebra, especially as exact values are preferred over decimal approximations.

To manipulate surds efficiently, it often involves rationalizing them. This process allows the conversion of surds in the denominator to a rational number, simplifying the expression:

• Multiply both the numerator and denominator by the conjugate of the denominator.
• This uses the identity \((a+b)(a-b) = a^2 - b^2\), eliminating the surd in the denominator.
• The result is a simpler form, easier to work with.

Following these procedures helps in computations and reveals simpler forms often hidden in more complicated expressions.

Simplification of Expressions

Simplification of expressions is a core skill in algebra where expressions are rewritten as simply as possible. It often involves reducing the complexity of algebraic expressions to make them easier to understand or solve. This could mean factoring, expanding, and combining like terms in an expression.

In the context of rationalizing denominators, simplification is used to make fractions easier to handle by removing surds from the denominator. The identities \((a+b)(a-b)=a^2-b^2\) play a key role here, as demonstrated in the multiplication of conjugates. For expression \(\frac{1}{a+b\sqrt{c}}\), we multiply by \(a-b\sqrt{c}\):

• This results in a difference of squares.
• The surd is eliminated from the denominator.
• The expression is simplified meaningfully.

This simplification process is powerful in making problematic expressions more accessible and manageable, allowing for the application of further algebraic operations.