Solve The Following Equation By Making An Appropriate Substitution:$x^{-2} + 8x^{-1} + 15 = 0$Select The Correct Choice Below And, If Necessary, Provide The Solution.

Introduction

In this article, we will be solving the equation $x^{-2} + 8x^{-1} + 15 = 0$ using an appropriate substitution. This type of equation is known as a quadratic equation in disguise, and we can solve it by making a substitution that will transform it into a quadratic equation.

Understanding the Equation

The given equation is $x^{-2} + 8x^{-1} + 15 = 0$. This equation can be rewritten as $\frac{1}{x^2} + \frac{8}{x} + 15 = 0$. We can see that the equation has a negative exponent, which can make it difficult to solve. To make it easier to solve, we can make a substitution that will eliminate the negative exponent.

Making a Substitution

Let's make a substitution by letting $y = \frac{1}{x}$. This means that $x = \frac{1}{y}$. We can now substitute this expression for $x$ into the original equation.

Substituting $y = \frac{1}{x}$ into the Equation

Substituting $y = \frac{1}{x}$ into the equation, we get:

$$\frac{1}{\left(\frac{1}{y}\right)^2} + \frac{8}{\frac{1}{y}} + 15 = 0$$

Simplifying this expression, we get:

$$y^2 + 8y + 15 = 0$$

Solving the Quadratic Equation

Now that we have a quadratic equation, we can solve it using the quadratic formula or factoring. Let's try factoring.

Factoring the Quadratic Equation

The quadratic equation $y^2 + 8y + 15 = 0$ can be factored as:

$$(y + 3)(y + 5) = 0$$

This means that either $y + 3 = 0$ or $y + 5 = 0$.

Solving for $y$

Solving for $y$, we get:

$$y + 3 = 0 \Rightarrow y = -3$$

$$y + 5 = 0 \Rightarrow y = -5$$

Substituting Back to Solve for $x$

Now that we have the values of $y$, we can substitute back to solve for $x$. Remember that $y = \frac{1}{x}$, so we can substitute $y = -3$ and $y = -5$ back into this equation.

Solving for $x$

Solving for $x$, we get:

$$\frac{1}{x} = -3 \Rightarrow x = -\frac{1}{3}$$

$$\frac{1}{x} = -5 \Rightarrow x = -\frac{1}{5}$$

Conclusion

In this article, we solved the equation $x^{-2} + 8x^{-1} + 15 = 0$ using an appropriate substitution. We made a substitution by letting $y = \frac{1}{x}$, which transformed the equation into a quadratic equation. We then solved the quadratic equation using factoring and substituted back to solve for $x$. The solutions to the equation are $x = -\frac{1}{3}$ and $x = -\frac{1}{5}$.

Final Answer

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Introduction

In our previous article, we solved the equation $x^{-2} + 8x^{-1} + 15 = 0$ using an appropriate substitution. This type of equation is known as a quadratic equation in disguise, and we can solve it by making a substitution that will transform it into a quadratic equation. In this article, we will answer some frequently asked questions about solving equations with an appropriate substitution.

Q&A

Q: What is an appropriate substitution?

A: An appropriate substitution is a substitution that will transform the equation into a quadratic equation. In the case of the equation $x^{-2} + 8x^{-1} + 15 = 0$, we made a substitution by letting $y = \frac{1}{x}$.

Q: Why do we need to make a substitution?

A: We need to make a substitution because the equation has a negative exponent, which can make it difficult to solve. By making a substitution, we can eliminate the negative exponent and transform the equation into a quadratic equation.

Q: How do we know which substitution to make?

A: We can determine which substitution to make by looking at the equation and identifying the negative exponent. In this case, we had $x^{-2}$, so we made a substitution by letting $y = \frac{1}{x}$.

Q: Can we use any substitution we want?

A: No, we cannot use any substitution we want. The substitution must be an appropriate one that will transform the equation into a quadratic equation. If we make a substitution that does not work, we will not be able to solve the equation.

Q: How do we solve the quadratic equation?

A: We can solve the quadratic equation using the quadratic formula or factoring. In this case, we factored the quadratic equation $y^2 + 8y + 15 = 0$ as $(y + 3)(y + 5) = 0$.

Q: What if the quadratic equation does not factor easily?

A: If the quadratic equation does not factor easily, we can use the quadratic formula to solve it. The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation.

Q: Can we use the quadratic formula with a negative exponent?

A: No, we cannot use the quadratic formula with a negative exponent. The quadratic formula is used to solve quadratic equations, and it does not work with negative exponents.

Q: What if we make a mistake and get an incorrect solution?

A: If we make a mistake and get an incorrect solution, we can try again and make sure to double-check our work. We can also use a calculator or computer program to check our solution.

Q: Can we use an appropriate substitution with any type of equation?

A: No, we cannot use an appropriate substitution with any type of equation. The substitution must be an appropriate one that will transform the equation into a quadratic equation. If we try to use an appropriate substitution with an equation that is not quadratic, we will not be able to solve it.

Conclusion

In this article, we answered some frequently asked questions about solving equations with an appropriate substitution. We discussed the importance of making an appropriate substitution, how to determine which substitution to make, and how to solve the quadratic equation. We also discussed some common mistakes that people make when solving equations with an appropriate substitution.

Final Answer

The final answer is: There is no final numerical answer to this problem. The article is a Q&A about solving equations with an appropriate substitution.