Problem 44 - Algebra 1

Question

Simplify the expression. $$\frac{6}{10+\sqrt{2}}$$

Step-by-Step Solution

Answer

The simplified expression is: \(\frac{30 - 3\sqrt{2}}{49}\)

Step 1: Recognize what needs to be simplified

We can see that the denominator of the fraction contains the square root of a number. These types of expressions can often be simplified by rationalizing the denominator, or by changing it into a rational number. We do this by multiplying both the numerator and the denominator by the conjugate of the denominator.

Step 2: Multiply by the conjugate

The conjugate of \(10 + \sqrt{2}\) is \(10 - \sqrt{2}\). So we multiply both the numerator and the denominator by \(10 - \sqrt{2}\): \[\frac{6}{10+\sqrt{2}} \times \frac{10 - \sqrt{2}}{10 - \sqrt{2}}\] This simplifies to: \[\frac{6(10 - \sqrt{2})}{10^2 - (\sqrt{2})^2}\].

Step 3: Simplify the expression

When squared, \(10^2 = 100\) and \((\sqrt{2})^2 = 2\). So we have: \[\frac{6(10 - \sqrt{2})}{100 - 2}\] which simplifies to: \[\frac{60 - 6\sqrt{2}}{98}\]

Step 4: Simplify further

We can further simplify the expression by factoring out a 2 from the denominator: \[\frac{60 - 6\sqrt{2}}{98} = \frac{30 - 3\sqrt{2}}{49}\]

Key Concepts

Simplify Expressions

Simplifying mathematical expressions is a fundamental skill in algebra. It involves altering the form of an expression without changing its value. This results in a more understandable and often more useful expression. Simplifying can include actions such as combining like terms, reducing fractions, and removing radical terms from denominators.

In the provided exercise, the expression \( \frac{6}{10+\sqrt{2}} \) needs simplification, specifically targeting the denominator containing a square root. Simplifying such expressions often leads to clearer understanding and easier manipulation, especially when dealing with complex equations later on.

Conjugate of a Number

In algebra, the conjugate of a number refers to a binomial formed by changing the sign of only one term in the binomial. It is particularly useful when dealing with square roots. The conjugate of the binomial \(a + \sqrt{b}\) is \(a - \sqrt{b}\).

Why Use A Conjugate?

Conjugates can be used to rationalize denominators containing square roots—such as in the exercise where the conjugate of \(10 + \sqrt{2}\) is \(10 - \sqrt{2}\). Multiplying a fraction by the conjugate of its denominator does not change its value but creates a difference of two squares in the denominator. This is crucial because it eliminates the square root, making the expression easier to work with.

Square Roots

Square roots are essential in algebra, providing a solution to the question 'What number, when multiplied by itself, gives the original number?' For example, \(\sqrt{9} = 3\) because \(3 \times 3 = 9\). Simplifying square roots often involves identifying and factoring out perfect squares, reducing the expression to its simplest radical form.

In the context of rationalization, the square root plays a key role, as seen in the exercise. The square of \(\sqrt{2}\) is \(2\), which becomes part of the 'difference of squares' when we multiply by the conjugate. This allows us to further simplify the expression, as the intimidating radical is no more in the denominator.