Express 8/ square root 5 - 1 in the form square root a + b where a and b are integers 

To express \(\frac{8}{\sqrt{5} - 1}\) in the form \(\sqrt{a} + b\), we'll follow these steps:

### Step 1: Multiply by the conjugate
To eliminate the square root from the denominator, multiply both the numerator and the denominator by the conjugate of \(\sqrt{5} - 1\), which is \(\sqrt{5} + 1\).

\[
\frac{8}{\sqrt{5} - 1} \times \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{8(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)}
\]

### Step 2: Simplify the denominator
The denominator is a difference of squares:

\[
(\sqrt{5} - 1)(\sqrt{5} + 1) = \sqrt{5}^2 - 1^2 = 5 - 1 = 4
\]

### Step 3: Simplify the expression
Now simplify the numerator:

\[
\frac{8(\sqrt{5} + 1)}{4} = \frac{8\sqrt{5} + 8}{4}
\]

Now divide both terms in the numerator by 4:

\[
= 2\sqrt{5} + 2
\]

### Final Answer:
Thus, \(\frac{8}{\sqrt{5} - 1}\) can be expressed as \(2\sqrt{5} + 2\).

---

Would you like further details or have any questions? Here are some related problems you might find useful:

1. Simplify \(\frac{10}{\sqrt{7} - 2}\) into the form \(\sqrt{a} + b\).
2. Express \(\frac{12}{\sqrt{3} + 2}\) in the form \(\sqrt{a} + b\).
3. Simplify \(\frac{15}{\sqrt{2} - 3}\) in terms of \(\sqrt{a} + b\).
4. Find the value of \(\frac{20}{\sqrt{6} + 1}\) in simplest form.
5. Express \(\frac{9}{2 + \sqrt{11}}\) as \(\sqrt{a} + b\).

**Tip:** When simplifying expressions with square roots in the denominator, multiplying by the conjugate is a key technique to eliminate the square root.

Express 8/ square root 5 - 1 in the form square root a + b where a and b are integers.

This guide walks through simplifying the expression 8 divided by (√5 - 1) into the form √a + b, where a and b are integers. Using the conjugate multiplication technique and the difference of squares formula, the expression is simplified to 2√5 + 2. Ideal for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation