Use The Following Table To Find The Value Of { X $}$ That Satisfies The Equation { \frac{1}{x}+4=x^2+3$} . . . [ \begin{tabular}{|c|c|c|} \hline X X X & F ( X ) = 1 X + 4 F(x)=\frac{1}{x}+4 F ( X ) = X 1 ​ + 4 & G ( X ) = X 2 + 3 G(x)=x^2+3 G ( X ) = X 2 + 3 \ \hline 1.0 & 5.00 & 4.00 \ \hline 1.1 &

Introduction

In this article, we will explore a mathematical equation and use a table to find the value of $x$ that satisfies the equation. The equation is given as $\frac{1}{x}+4=x^2+3$. We will use a table to help us find the solution.

Understanding the Equation

The equation $\frac{1}{x}+4=x^2+3$ is a quadratic equation. It can be rewritten as $x^2 - \frac{1}{x} - 1 = 0$. This equation is a quadratic equation in terms of $x$, and we need to find the value of $x$ that satisfies this equation.

Using a Table to Find the Solution

We are given a table with three columns: $x$, $f(x)=\frac{1}{x}+4$, and $g(x)=x^2+3$. We can use this table to help us find the solution to the equation.

$x$	$f(x)=\frac{1}{x}+4$	$g(x)=x^2+3$
1.0	5.00	4.00
1.1	5.09	4.21

Analyzing the Table

From the table, we can see that the values of $f(x)$ and $g(x)$ are close to each other for $x=1.1$. This suggests that the solution to the equation may be close to $x=1.1$.

Finding the Solution

To find the solution, we need to find the value of $x$ that satisfies the equation $x^2 - \frac{1}{x} - 1 = 0$. We can use the table to help us find this value.

Let's start by looking at the values of $f(x)$ and $g(x)$ for $x=1.1$. We can see that $f(1.1)=5.09$ and $g(1.1)=4.21$. Since $f(1.1)$ is greater than $g(1.1)$, we know that the solution to the equation must be greater than $x=1.1$.

Using the Table to Find the Solution

We can use the table to find the solution by looking at the values of $f(x)$ and $g(x)$ for different values of $x$. We can see that the values of $f(x)$ and $g(x)$ are close to each other for $x=1.1$, and the solution to the equation must be greater than $x=1.1$.

Finding the Solution Using Algebra

We can also use algebra to find the solution to the equation. We can start by rewriting the equation as $x^2 - \frac{1}{x} - 1 = 0$. We can then multiply both sides of the equation by $x$ to get $x^3 - 1 - x = 0$. We can then factor the left-hand side of the equation to get $(x-1)(x^2+x+1)=0$. We can then set each factor equal to zero and solve for $x$.

Solving for $x$

We can set the first factor equal to zero and solve for $x$ to get $x-1=0$, which gives us $x=1$. We can set the second factor equal to zero and solve for $x$ to get $x^2+x+1=0$. We can use the quadratic formula to solve for $x$ to get $x=\frac{-1\pm\sqrt{1-4(1)(1)}}{2(1)}=\frac{-1\pm\sqrt{-3}}{2}$. Since the square root of a negative number is not a real number, we know that the solution to the equation must be $x=1$.

Conclusion

In this article, we used a table to find the value of $x$ that satisfies the equation $\frac{1}{x}+4=x^2+3$. We also used algebra to find the solution to the equation. We found that the solution to the equation is $x=1$.

References

• [1] "Algebra" by Michael Artin
• [2] "Calculus" by Michael Spivak

Appendix

The following is a list of the values of $f(x)$ and $g(x)$ for different values of $x$.

$x$	$f(x)=\frac{1}{x}+4$	$g(x)=x^2+3$
1.0	5.00	4.00
1.1	5.09	4.21
1.2	5.18	4.44
1.3	5.27	4.69
1.4	5.36	4.96
1.5	5.45	5.25
1.6	5.54	5.56
1.7	5.63	5.89
1.8	5.72	6.24
1.9	5.81	6.61
2.0	5.90	7.00

$$$$

Q: What is the equation that we are trying to solve?

A: The equation that we are trying to solve is $\frac{1}{x}+4=x^2+3$.

Q: How do we use the table to find the solution to the equation?

A: We can use the table to find the solution to the equation by looking at the values of $f(x)$ and $g(x)$ for different values of $x$. We can see that the values of $f(x)$ and $g(x)$ are close to each other for $x=1.1$, and the solution to the equation must be greater than $x=1.1$.

Q: How do we find the solution to the equation using algebra?

A: We can use algebra to find the solution to the equation by rewriting the equation as $x^2 - \frac{1}{x} - 1 = 0$. We can then multiply both sides of the equation by $x$ to get $x^3 - 1 - x = 0$. We can then factor the left-hand side of the equation to get $(x-1)(x^2+x+1)=0$. We can then set each factor equal to zero and solve for $x$.

Q: What is the solution to the equation?

A: The solution to the equation is $x=1$.

Q: Why is the solution to the equation $x=1$?

A: The solution to the equation is $x=1$ because when we substitute $x=1$ into the equation, we get $\frac{1}{1}+4=1^2+3$, which simplifies to $5=4$, which is a true statement.

Q: What if we want to find the solution to the equation using a different method?

A: If we want to find the solution to the equation using a different method, we can try using a different algebraic technique or a numerical method. However, in this case, we found that the solution to the equation is $x=1$ using the method of factoring.

Q: Can we use the table to find the solution to other equations?

A: Yes, we can use the table to find the solution to other equations. However, we need to make sure that the equation is in the form of $f(x)=g(x)$, where $f(x)$ and $g(x)$ are functions of $x$.

Q: How do we create a table to find the solution to an equation?

A: To create a table to find the solution to an equation, we need to choose a range of values for $x$ and calculate the corresponding values of $f(x)$ and $g(x)$. We can then use the table to find the solution to the equation by looking at the values of $f(x)$ and $g(x)$ for different values of $x$.

Q: What are some common mistakes to avoid when using a table to find the solution to an equation?

A: Some common mistakes to avoid when using a table to find the solution to an equation include:

• Not choosing a range of values for $x$ that is large enough to capture the solution to the equation.
• Not calculating the corresponding values of $f(x)$ and $g(x)$ accurately.
• Not using the table to find the solution to the equation correctly.

Q: How do we know if the solution to the equation is accurate?

A: We can know if the solution to the equation is accurate by checking our work and making sure that we have not made any mistakes. We can also use a calculator or computer software to check our solution and make sure that it is accurate.

Q: Can we use the table to find the solution to a system of equations?

A: Yes, we can use the table to find the solution to a system of equations. However, we need to make sure that the system of equations is in the form of $f(x)=g(x)$ and $h(x)=k(x)$, where $f(x)$, $g(x)$, $h(x)$, and $k(x)$ are functions of $x$.

Q: How do we create a table to find the solution to a system of equations?

A: To create a table to find the solution to a system of equations, we need to choose a range of values for $x$ and calculate the corresponding values of $f(x)$, $g(x)$, $h(x)$, and $k(x)$. We can then use the table to find the solution to the system of equations by looking at the values of $f(x)$, $g(x)$, $h(x)$, and $k(x)$ for different values of $x$.