Problem 46 - Algebra 1

Question

Simplify the expression. $$\frac{14}{60-\sqrt{578}}$$

Step-by-Step Solution

Answer

The simplified form of \(\frac{14}{60 - \sqrt{578}}\) is \(\frac{60 + \sqrt{578}}{216}\)

Step 1: Expression Rationalization

Rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator. The conjugate of \(60 - \sqrt{578}\) is \(60 + \sqrt{578}\). So, we multiply the fraction by \(\frac{60 + \sqrt{578}}{60 + \sqrt{578}}\), which gives us \(\frac{14 \times (60 + \sqrt{578})}{(60 - \sqrt{578})*(60 + \sqrt{578})}\)

Step 2: Simplify The Denominator Using Difference of Squares

The denominator is a difference of squares, which is a formula that can be simplified to \((a-b)(a+b) = a^2 - b^2\). Applying that template to the denominator, we obtain: \(\frac{14 \times (60 + \sqrt{578})}{60^2 - (\sqrt{578}^2}\) which simplifies to \(\frac{14 \times (60 + \sqrt{578})}{3600 - 578}\)

Step 3: Simplify The Denominator

Subtract 578 from 3600 in the denominator, which results in \(\frac{14 \times (60 + \sqrt{578})}{3022}\)

Step 4: Distribute the Numerator

Multiply \(14 \times 60, 14 \times \sqrt{578}\) to simplify the numerator, yielding \(\frac{840 + 14\sqrt{578}}{3022}\)

Step 5: Simplify The Whole Expression

Divide all terms by the greatest common divisor of 840, 14, and 3022 which is 14, yielding \(\frac{60 + \sqrt{578}}{216}\)

Key Concepts

Rationalizing the Denominator

When faced with a radical expression that has a square root in the denominator, it is often necessary to "rationalize the denominator". This means transforming the expression so the denominator becomes a rational number, free from radicals.

In the exercise, the original expression is \(\frac{14}{60-\sqrt{578}}\). To rationalize the denominator here, we multiply both the numerator and the denominator by the conjugate of the denominator. A conjugate involves changing the sign between terms in a binomial. Thus, the conjugate of \(60 - \sqrt{578}\) is \(60 + \sqrt{578}\). By multiplying the entire fraction by \(\frac{60 + \sqrt{578}}{60 + \sqrt{578}}\), we take advantage of the property that allows the elimination of the radical in the denominator.

This approach creates a difference of squares, which makes the denominator a simple difference between squares of real numbers. This results in a rational number denominator, simplifying our calculations greatly.

Difference of Squares

In mathematics, the "difference of squares" is a specific type of polynomial factoring. The formula is \((a-b)(a+b) = a^2 - b^2\). This is incredibly useful when rationalizing expressions with radicals.

Applying it to the exercise, the denominator becomes a difference of squares: \((60-\sqrt{578})(60+\sqrt{578})\). The expression simplifies to \(60^2 - (\sqrt{578})^2\).

Calculations using this formula transform complex expressions into simpler ones. Here, \(60^2\) results in 3600, and \((\sqrt{578})^2\) equals 578. By performing the subtraction, we remove the radical completely, arriving at the denominator 3022, which is completely devoid of radicals. This method is efficient and eliminates the need for radical expressions in denominators, making further simplification easier.

Simplifying Expressions

"Simplifying Expressions" is about breaking down complex mathematical expressions into their simplest, most manageable forms. It involves reducing fractions, combining like terms, and using arithmetic operations to decrease complexity while retaining the original value.

In our exercise, we start with a complex fraction expression \(\frac{14 \times (60 + \sqrt{578})}{3022}\). First, distribute 14 across the terms within the parentheses, resulting in a numerator of \(840 + 14\sqrt{578}\).

This step-by-step process of simplification reduces the expression to \(\frac{60 + \sqrt{578}}{216}\). The greatest common factor, 14, is factored out from each numeric component, simplifying the expression to its most reduced form. This simplification makes calculations easier, minimizes errors, and is crucial for solving mathematical problems efficiently.