Problem 71 - Technical Mathematics with Calculus

Question

Challenge Problems. Perform the indicated operation and simplify. $$\sqrt{x} \div(\sqrt{x}+\sqrt{y})$$

Step-by-Step Solution

Answer

The simplified expression is \(\frac{x - \sqrt{xy}}{x - y}\).

Step 1: Simplify the expression

To simplify the expression \(\sqrt{x} \div (\sqrt{x}+\sqrt{y})\), you can rewrite it as \(\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}\).

Step 2: Rationalize the denominator

Multiply the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{x}-\sqrt{y}\), in order to get rid of the square roots in the denominator.

Step 3: Perform the multiplication

Multiply \(\sqrt{x}\) by \(\sqrt{x}-\sqrt{y}\) to get \(x - \sqrt{xy}\) in the numerator. Also, multiply \(\sqrt{x}+\sqrt{y}\) by \(\sqrt{x}-\sqrt{y}\) in the denominator which simplifies to \(x - y\) because it is a difference of squares.

Step 4: Rewrite simplified expression

The simplified expression after rationalization is \(\frac{x - \sqrt{xy}}{x - y}\).

Key Concepts

Rationalize the Denominator

When we come across a radical in the denominator of an expression, a common practice is to rationalize that denominator. This means we want to eliminate any roots or radicals from the bottom part of a fraction, creating a 'nicer' expression for further mathematical operations or simply for a final, more aesthetically pleasing result.

To rationalize the denominator, we multiply both the top and the bottom of the fraction by the conjugate of the denominator if it's a binomial involving square roots. The conjugate is similar to the original expression, but with the opposite sign between its terms. In our specific problem, we use \(\sqrt{x} - \sqrt{y}\) to rationalize \(\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}\). This method avoids changing the value of the fraction since, effectively, we are multiplying by one in a disguised form.

Difference of Squares

The difference of squares is a particular algebraic pattern you will often encounter: \(a^2 - b^2\). This pattern is remarkable because it can be factored into \(a + b\) times \(a - b\). When we multiply the conjugate of our denominator by the original denominator in the problem \(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y}\), we get the perfect example of the difference of squares because \(\sqrt{x}\) and \(\sqrt{y}\) are like \(a\) and \(b\), respectively.

The beauty of using this method in rationalization is that the square roots cancel out, leaving us with whole numbers or expressions free of radicals. For instance, when we multiplied \(\sqrt{x}+\sqrt{y}\) by its conjugate, the result simplified to \(x - y\), since by the pattern \(\sqrt{x}^2\) simplifies to \(x\) and \(\sqrt{y}^2\) to \(y\). This leaves our expression in a cleaner and more workable form, especially when solving equations or simplifying further.

Simplification of Square Roots

Simplifying square roots, also known as radicals, is an essential skill to make complex expressions manageable. To simplify a square root, we look for the largest square factor of the number under the radical sign and break it down. For example, if we have \(\sqrt{18}\), we would recognize that 18 = 9 * 2, and since 9 is a perfect square, we can simplify \(\sqrt{18}\) to \(3\sqrt{2}\).

In the context of our problem, simplification comes after we have rationalized the denominator and performed the difference of squares. After these steps, the square roots are either eliminated from the denominator or reduced to simpler forms in the numerator, thus making the overall expression easier to work with in further mathematics.