7. If $a = 4 - \sqrt{15}$, Find The Value Of:1. $a^2 + \frac{1}{a^2}$2. $a^3 + \frac{1}{a^3}$

Introduction

In this article, we will explore the process of solving equations involving square roots and cubic roots. We will use the given equation $a = 4 - \sqrt{15}$ to find the value of two different expressions: $a^2 + \frac{1}{a^2}$ and $a^3 + \frac{1}{a^3}$. These types of problems often require the use of algebraic manipulations and the application of mathematical identities.

Finding the Value of $a^2 + \frac{1}{a^2}$

To find the value of $a^2 + \frac{1}{a^2}$, we can start by squaring the given equation $a = 4 - \sqrt{15}$.

Squaring the Equation

$a^2 = (4 - \sqrt{15})^2$

Using the formula $(a - b)^2 = a^2 - 2ab + b^2$, we can expand the equation as follows:

$a^2 = 4^2 - 2 \cdot 4 \cdot \sqrt{15} + (\sqrt{15})^2$

$a^2 = 16 - 8\sqrt{15} + 15$

$a^2 = 31 - 8\sqrt{15}$

Finding the Value of $\frac{1}{a^2}$

To find the value of $\frac{1}{a^2}$, we can take the reciprocal of the equation $a^2 = 31 - 8\sqrt{15}$.

$\frac{1}{a^2} = \frac{1}{31 - 8\sqrt{15}}$

To simplify this expression, we can multiply the numerator and denominator by the conjugate of the denominator, which is $31 + 8\sqrt{15}$.

$\frac{1}{a^2} = \frac{31 + 8\sqrt{15}}{(31 - 8\sqrt{15})(31 + 8\sqrt{15})}$

Using the formula $(a - b)(a + b) = a^2 - b^2$, we can simplify the denominator as follows:

$\frac{1}{a^2} = \frac{31 + 8\sqrt{15}}{31^2 - (8\sqrt{15})^2}$

$\frac{1}{a^2} = \frac{31 + 8\sqrt{15}}{961 - 720}$

$\frac{1}{a^2} = \frac{31 + 8\sqrt{15}}{241}$

Finding the Value of $a^2 + \frac{1}{a^2}$

Now that we have found the values of $a^2$ and $\frac{1}{a^2}$, we can add them together to find the value of $a^2 + \frac{1}{a^2}$.

$a^2 + \frac{1}{a^2} = (31 - 8\sqrt{15}) + \frac{31 + 8\sqrt{15}}{241}$

$a^2 + \frac{1}{a^2} = \frac{241(31 - 8\sqrt{15}) + 31 + 8\sqrt{15}}{241}$

$a^2 + \frac{1}{a^2} = \frac{7471 - 1944\sqrt{15} + 31 + 8\sqrt{15}}{241}$

$a^2 + \frac{1}{a^2} = \frac{7502 - 1936\sqrt{15}}{241}$

Finding the Value of $a^3 + \frac{1}{a^3}$

To find the value of $a^3 + \frac{1}{a^3}$, we can start by cubing the given equation $a = 4 - \sqrt{15}$.

Cubing the Equation

$a^3 = (4 - \sqrt{15})^3$

Using the formula $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$, we can expand the equation as follows:

$a^3 = 4^3 - 3 \cdot 4^2 \cdot \sqrt{15} + 3 \cdot 4 \cdot (\sqrt{15})^2 - (\sqrt{15})^3$

$a^3 = 64 - 48\sqrt{15} + 180 - 15\sqrt{15}$

$a^3 = 244 - 63\sqrt{15}$

Finding the Value of $\frac{1}{a^3}$

To find the value of $\frac{1}{a^3}$, we can take the reciprocal of the equation $a^3 = 244 - 63\sqrt{15}$.

$\frac{1}{a^3} = \frac{1}{244 - 63\sqrt{15}}$

To simplify this expression, we can multiply the numerator and denominator by the conjugate of the denominator, which is $244 + 63\sqrt{15}$.

$\frac{1}{a^3} = \frac{244 + 63\sqrt{15}}{(244 - 63\sqrt{15})(244 + 63\sqrt{15})}$

Using the formula $(a - b)(a + b) = a^2 - b^2$, we can simplify the denominator as follows:

$\frac{1}{a^3} = \frac{244 + 63\sqrt{15}}{244^2 - (63\sqrt{15})^2}$

$\frac{1}{a^3} = \frac{244 + 63\sqrt{15}}{59536 - 47835}$

$\frac{1}{a^3} = \frac{244 + 63\sqrt{15}}{11701}$

Finding the Value of $a^3 + \frac{1}{a^3}$

Now that we have found the values of $a^3$ and $\frac{1}{a^3}$, we can add them together to find the value of $a^3 + \frac{1}{a^3}$.

$a^3 + \frac{1}{a^3} = (244 - 63\sqrt{15}) + \frac{244 + 63\sqrt{15}}{11701}$

$a^3 + \frac{1}{a^3} = \frac{11701(244 - 63\sqrt{15}) + 244 + 63\sqrt{15}}{11701}$

$a^3 + \frac{1}{a^3} = \frac{2833044 - 7371075\sqrt{15} + 244 + 63\sqrt{15}}{11701}$

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Q: What is the main difference between solving equations involving square roots and cubic roots?

A: The main difference between solving equations involving square roots and cubic roots is the complexity of the equations. Equations involving square roots can be solved using the formula $(a - b)^2 = a^2 - 2ab + b^2$, while equations involving cubic roots require the use of the formula $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$.

Q: How do I simplify expressions involving square roots and cubic roots?

A: To simplify expressions involving square roots and cubic roots, you can use the conjugate of the denominator to eliminate the radical. For example, if you have the expression $\frac{1}{a^2}$, you can multiply the numerator and denominator by the conjugate of the denominator, which is $a^2 + \frac{1}{a^2}$.

Q: What is the formula for finding the value of $a^2 + \frac{1}{a^2}$?

A: The formula for finding the value of $a^2 + \frac{1}{a^2}$ is:

$a^2 + \frac{1}{a^2} = (a^2 + \frac{1}{a^2}) \cdot \frac{a^2 - \frac{1}{a^2}}{a^2 - \frac{1}{a^2}}$

This can be simplified to:

$a^2 + \frac{1}{a^2} = \frac{(a^2)^2 - (\frac{1}{a^2})^2}{a^2 - \frac{1}{a^2}}$

Q: What is the formula for finding the value of $a^3 + \frac{1}{a^3}$?

A: The formula for finding the value of $a^3 + \frac{1}{a^3}$ is:

$a^3 + \frac{1}{a^3} = (a^3 + \frac{1}{a^3}) \cdot \frac{a^3 - \frac{1}{a^3}}{a^3 - \frac{1}{a^3}}$

This can be simplified to:

$a^3 + \frac{1}{a^3} = \frac{(a^3)^2 - (\frac{1}{a^3})^2}{a^3 - \frac{1}{a^3}}$

Q: How do I apply the formulas for finding the value of $a^2 + \frac{1}{a^2}$ and $a^3 + \frac{1}{a^3}$?

A: To apply the formulas for finding the value of $a^2 + \frac{1}{a^2}$ and $a^3 + \frac{1}{a^3}$, you need to substitute the given values of $a$ into the formulas and simplify the expressions.

For example, if you have the equation $a = 4 - \sqrt{15}$, you can substitute this value into the formula for finding the value of $a^2 + \frac{1}{a^2}$:

$a^2 + \frac{1}{a^2} = (31 - 8\sqrt{15}) + \frac{31 + 8\sqrt{15}}{241}$

You can then simplify this expression to find the value of $a^2 + \frac{1}{a^2}$.

Q: What are some common mistakes to avoid when solving equations involving square roots and cubic roots?

A: Some common mistakes to avoid when solving equations involving square roots and cubic roots include:

• Not simplifying the expressions correctly
• Not using the conjugate of the denominator to eliminate the radical
• Not applying the formulas correctly
• Not checking the solutions for validity

By avoiding these common mistakes, you can ensure that you are solving the equations correctly and obtaining the correct solutions.