Problem 51
Question
Rationalize the denominator. $$ \frac{7}{\sqrt{5}-2} $$
Step-by-Step Solution
Verified Answer
The denominator is rationalized to give a final answer of \(7\sqrt{5} + 14\).
1Step 1: Identify the Conjugate
We identify the conjugate of the denominator \(\sqrt{5} - 2\). It is \(\sqrt{5} + 2\).
2Step 2: Multiply by Conjugate
We multiply both the numerator and the denominator of the fraction by this conjugate to rationalize it. Doing so gives: \(\frac{7 * (\sqrt{5} + 2)}{(\sqrt{5} - 2) * (\sqrt{5} + 2)}\).
3Step 3: Simplify
Using the identity \(a^2 - b^2 = (a+b)(a-b)\), we simplify the denominator to obtain \(5 - 4 = 1\). And the numerator is \(7\sqrt{5} + 14\). Putting it all together, the final expression is \(\frac{7\sqrt{5} + 14}{1}\), which simplifies to \(7\sqrt{5} + 14\).
Key Concepts
Conjugate of a NumberSimplify ExpressionsDifference of Squares Identity
Conjugate of a Number
The concept of a conjugate is crucial when dealing with complex numbers or expressions involving square roots. A conjugate in mathematics is formed by changing the sign between two terms. If you have a binomial expression like \( a + b \), the conjugate would be \( a - b \), and vice versa. This approach is particularly useful in rationalizing denominators.
In the context of our original exercise with the expression \( \frac{7}{\sqrt{5}-2} \), the denominator \( \sqrt{5}-2 \) has a conjugate of \( \sqrt{5}+2 \). Multiplying the original expression by the conjugate over itself \( \frac{\sqrt{5}+2}{\sqrt{5}+2} \) is a nifty trick. This step does not change the expression's value as we're essentially multiplying by 1, but it does help in removing the radical from the denominator.
In the context of our original exercise with the expression \( \frac{7}{\sqrt{5}-2} \), the denominator \( \sqrt{5}-2 \) has a conjugate of \( \sqrt{5}+2 \). Multiplying the original expression by the conjugate over itself \( \frac{\sqrt{5}+2}{\sqrt{5}+2} \) is a nifty trick. This step does not change the expression's value as we're essentially multiplying by 1, but it does help in removing the radical from the denominator.
Simplify Expressions
To simplify expressions, especially when involving radicals or complex fractions, you follow certain algebraic rules to make the expression easier to understand and work with. The goal is to make the numbers and terms as 'simple' as possible, which often means getting rid of radicals in the denominator or combining like terms.
For our example, once we have multiplied by the conjugate, we simplify the numerator and denominator separately. In the numerator, multiply out \( 7(\sqrt{5}+2) \) to get \( 7\sqrt{5} + 14 \). Simplifying the denominator involves recognizing a standard identity known as the difference of squares, which we’ll look at in more detail in the next section. After simplification, the expression no longer has a radical in the denominator and reads as a simple sum: \( 7\sqrt{5} + 14 \). Such simplification is a common technique and important for ensuring clarity and ease in further mathematical operations like addition, subtraction, or comparison.
For our example, once we have multiplied by the conjugate, we simplify the numerator and denominator separately. In the numerator, multiply out \( 7(\sqrt{5}+2) \) to get \( 7\sqrt{5} + 14 \). Simplifying the denominator involves recognizing a standard identity known as the difference of squares, which we’ll look at in more detail in the next section. After simplification, the expression no longer has a radical in the denominator and reads as a simple sum: \( 7\sqrt{5} + 14 \). Such simplification is a common technique and important for ensuring clarity and ease in further mathematical operations like addition, subtraction, or comparison.
Difference of Squares Identity
The difference of squares is an algebraic identity that states that for any two terms, \( a \) and \( b \), the product of their sum and their difference equals the difference of their squares: \[ a^2 - b^2 = (a+b)(a-b) \]. This identity is immensely useful in various areas of mathematics including simplifying algebraic expressions and solving equations.
In the exercise, after multiplying the denominator by the conjugate, we recognize the pattern that fits the difference of squares identity: \((\sqrt{5}-2)(\sqrt{5}+2)\). This simplifies to \(5 - 4\), because \( \sqrt{5}^2 = 5 \) and \( 2^2 = 4 \), ultimately leaving us with \(1\) as the denominator. At this point, the denominator is completely rationalized and the expression has been greatly simplified, demonstrating the power and utility of the difference of squares identity in algebra.
In the exercise, after multiplying the denominator by the conjugate, we recognize the pattern that fits the difference of squares identity: \((\sqrt{5}-2)(\sqrt{5}+2)\). This simplifies to \(5 - 4\), because \( \sqrt{5}^2 = 5 \) and \( 2^2 = 4 \), ultimately leaving us with \(1\) as the denominator. At this point, the denominator is completely rationalized and the expression has been greatly simplified, demonstrating the power and utility of the difference of squares identity in algebra.
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