Question

\(\frac{\sqrt{6}}{\sqrt{3}-\sqrt{5}}\)

Short answer

The simplified form of the expression \(\frac{\sqrt{6}}{\sqrt{3}-\sqrt{5}}\) is \(-3\sqrt{2} - \sqrt{30}\)

Step-by-step solution

Multiply by the Conjugate

Multiply the expression by \(\frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}}\), the conjugate of \(\sqrt{3}-\sqrt{5}\) to eliminate the square roots in the denominator. This results in: \(\frac{\sqrt{6}(\sqrt{3}+\sqrt{5})}{(\sqrt{3}-\sqrt{5})(\sqrt{3}+\sqrt{5})}\)

Simplify the Denominator

Simplify the term in the denominator. As it is the product of two expressions in the form \(a-b)(a+b)\), which is equal to \(a^2 - b^2\), we can apply this formula to our denominator. \(\frac{\sqrt{6}(\sqrt{3}+\sqrt{5})}{(\sqrt{3})^2 - (\sqrt{5})^2}\)

Simplify the Expression

Now simplify both the numerator and denominator, to obtain the final simplified form. \(\frac{\sqrt{18} + \sqrt{30}}{3 - 5} = \frac{\sqrt{18} + \sqrt{30}}{-2} \). Remember that the square root of a multiplication is the multiplication of the square roots, and that you can simplify the square root of 18 to 3*sqrt(2) and the square root of 30 to sqrt(2)*sqrt(15). This leads to \(-\sqrt{2}*\sqrt{9} - \sqrt{2}*\sqrt{15}\) = \(-3\sqrt{2} - \sqrt{30} \)

Key concepts

Conjugate Multiplication

Sometimes when dealing with radical expressions in the denominator, it is helpful to use a technique called conjugate multiplication. The conjugate of a binomial is just the same two terms but with the opposite middle sign. For example, for the expression \(\sqrt{3} - \sqrt{5}\), the conjugate is \(\sqrt{3} + \sqrt{5}\). The beauty of this technique is that it helps us eliminate radicals or surds in the denominator.

When we multiply a fraction by this conjugate over itself, we effectively change the denominator to a simpler form without changing the value of the expression, since any number divided by itself is 1. This helps to transform the denominator into a rational number by a method known as the difference of squares. It's an essential trick to "clear" or "rationalize" the denominator.

Simplifying Radical Expressions

Simplifying radical expressions often involves carrying out several key steps until you reach a form that can't be simplified further. At the heart of simplifying is reducing the radicand—the number inside the square root—to its simplest form.

One important step is to look for perfect squares within the radicand. For instance, the number 18 can be broken down into \(9\times2\), where 9 is a perfect square. This allows us to write \(\sqrt{18}\) as \(\sqrt{9} \times \sqrt{2}\), which simplifies to \(3\sqrt{2}\).

• Continuously break down the radicand into products of prime numbers or perfect squares when possible.
• Remember fundamental rules like \(\sqrt{a} \times \sqrt{b} = \sqrt{a \cdot b}\) which can simplify the manipulation of radicals.

Ultimately, the main goal of simplifying radicals is to end up with the most reduced version without any radicals in the denominator.

Difference of Squares

The difference of squares is an algebraic technique used to simplify expressions into a specific form. It's characterized by expressions that fit the format \(a^2 - b^2\), which simplifies conveniently to \((a-b)(a+b)\).

This concept plays a crucial role when used under the guise of conjugate multiplication. When we multiply \((\sqrt{3}-\sqrt{5})(\sqrt{3}+\sqrt{5})\), each root squared results in whole numbers. Specifically, \(\sqrt{3}^2\) equals 3 and \(\sqrt{5}^2\) equals 5, leading to the expression \(3-5\), which simplifies to -2.

• This method ensures that the radicals cancel out.
• It results in an integer or rational number that is much easier to work with.

Understanding and applying the difference of squares is fundamental in streamlining rationalization problems, making it a quintessential part of algebraic manipulation.