Factor And Solve The Following To Find All Roots:$\[ X^4 - 6x^2 - 7 = 0 \\] Degree: ?

Introduction

In this article, we will delve into the world of quartic equations and explore a method to factor and solve the given equation. The equation in question is $x^4 - 6x^2 - 7 = 0$. Our primary objective is to find all the roots of this equation and determine its degree.

Understanding Quartic Equations

A quartic equation is a polynomial equation of degree four, which means the highest power of the variable (in this case, $x$) is four. The general form of a quartic equation is $ax^4 + bx^3 + cx^2 + dx + e = 0$, where $a$, $b$, $c$, $d$, and $e$ are constants.

The Given Equation

Our given equation is $x^4 - 6x^2 - 7 = 0$. To begin solving this equation, we need to identify its degree. The degree of a polynomial is determined by the highest power of the variable. In this case, the highest power of $x$ is four, so the degree of the equation is four.

Factoring the Equation

To factor the equation, we can use a substitution method. Let's substitute $y = x^2$ into the equation. This will transform the equation into a quadratic equation in terms of $y$. The equation becomes $y^2 - 6y - 7 = 0$.

Solving the Quadratic Equation

Now that we have a quadratic equation in terms of $y$, we can solve it using the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. In this case, $a = 1$, $b = -6$, and $c = -7$. Plugging these values into the formula, we get:

$$y = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-7)}}{2(1)}$$

Simplifying the expression, we get:

$$y = \frac{6 \pm \sqrt{36 + 28}}{2}$$

$$y = \frac{6 \pm \sqrt{64}}{2}$$

$$y = \frac{6 \pm 8}{2}$$

This gives us two possible values for $y$: $y = \frac{6 + 8}{2} = 7$ and $y = \frac{6 - 8}{2} = -1$.

Substituting Back

Now that we have the values of $y$, we can substitute back to find the values of $x$. Remember, we substituted $y = x^2$. So, we have:

$$x^2 = 7 \quad \text{or} \quad x^2 = -1$$

Solving for $x$

To solve for $x$, we take the square root of both sides of the equation. For $x^2 = 7$, we get:

$$x = \pm \sqrt{7}$$

For $x^2 = -1$, we get:

$$x = \pm i\sqrt{1}$$

However, since $x^2 = -1$ has no real solutions, we discard this solution.

Conclusion

In conclusion, we have successfully factored and solved the quartic equation $x^4 - 6x^2 - 7 = 0$. The roots of the equation are $x = \pm \sqrt{7}$. The degree of the equation is four, which we determined at the beginning of our solution.

Final Answer

$$$$

Introduction

In our previous article, we explored the world of quartic equations and solved the equation $x^4 - 6x^2 - 7 = 0$. In this article, we will answer some frequently asked questions about quartic equations and provide additional insights into solving these types of equations.

Q: What is a quartic equation?

A: A quartic equation is a polynomial equation of degree four, which means the highest power of the variable (in this case, $x$) is four. The general form of a quartic equation is $ax^4 + bx^3 + cx^2 + dx + e = 0$, where $a$, $b$, $c$, $d$, and $e$ are constants.

Q: How do I determine the degree of a quartic equation?

A: To determine the degree of a quartic equation, you need to identify the highest power of the variable. In the case of the equation $x^4 - 6x^2 - 7 = 0$, the highest power of $x$ is four, so the degree of the equation is four.

Q: What is the difference between a quartic equation and a quadratic equation?

A: A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable is two. A quartic equation, on the other hand, is a polynomial equation of degree four, which means the highest power of the variable is four. This means that a quartic equation is more complex than a quadratic equation and requires different methods to solve.

Q: How do I factor a quartic equation?

A: To factor a quartic equation, you can use a substitution method. Let's say you have the equation $x^4 - 6x^2 - 7 = 0$. You can substitute $y = x^2$ into the equation, which will transform the equation into a quadratic equation in terms of $y$. You can then solve the quadratic equation using the quadratic formula.

Q: What is the quadratic formula?

A: The quadratic formula is a formula used to solve quadratic equations. It is given by $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. In the case of the equation $y^2 - 6y - 7 = 0$, $a = 1$, $b = -6$, and $c = -7$. Plugging these values into the formula, we get:

$$y = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-7)}}{2(1)}$$

Simplifying the expression, we get:

$$y = \frac{6 \pm \sqrt{36 + 28}}{2}$$

$$y = \frac{6 \pm \sqrt{64}}{2}$$

$$y = \frac{6 \pm 8}{2}$$

This gives us two possible values for $y$: $y = \frac{6 + 8}{2} = 7$ and $y = \frac{6 - 8}{2} = -1$.

Q: How do I substitute back to find the values of $x$?

A: To substitute back, you need to remember that you substituted $y = x^2$. So, you have:

$$x^2 = 7 \quad \text{or} \quad x^2 = -1$$

To solve for $x$, you take the square root of both sides of the equation. For $x^2 = 7$, you get:

$$x = \pm \sqrt{7}$$

For $x^2 = -1$, you get:

$$x = \pm i\sqrt{1}$$

However, since $x^2 = -1$ has no real solutions, you discard this solution.

Q: What are some common mistakes to avoid when solving quartic equations?

A: Some common mistakes to avoid when solving quartic equations include:

• Not identifying the degree of the equation correctly
• Not using the correct substitution method
• Not simplifying the expression correctly
• Not discarding complex solutions when they are not real

Conclusion

In conclusion, we have answered some frequently asked questions about quartic equations and provided additional insights into solving these types of equations. By following the steps outlined in this article, you should be able to solve quartic equations with ease.

Final Answer

The final answer is $\boxed{\sqrt{7}}$ and $\boxed{-\sqrt{7}}$.