How Can We Solve This Equation \left(1+x {2}\right) {1.5}=2+2xx
Introduction
In this article, we will delve into the world of linear algebra and explore a complex equation involving exponents and variables. The equation in question is \left(1+x{2}\right){1.5}=2+2xx. Our goal is to solve this equation and understand the underlying mathematical concepts that make it work.
Understanding the Equation
The given equation involves a variable x and an exponent of 1.5. To begin solving this equation, we need to understand the properties of exponents and how they interact with variables. The equation can be rewritten as \left(1+x{2}\right){1.5}=2+2xx.
Simplifying the Equation
To simplify the equation, we can start by isolating the variable x. We can do this by subtracting 2 from both sides of the equation, resulting in \left(1+x{2}\right){1.5}-2=2xx.
Using Algebraic Manipulation
Next, we can use algebraic manipulation to simplify the equation further. We can start by expanding the left-hand side of the equation using the binomial theorem. This will give us a polynomial expression in terms of x.
Expanding the Left-Hand Side
Using the binomial theorem, we can expand the left-hand side of the equation as follows:
\left(1+x{2}\right){1.5}=1+\frac{3}{2}x{2}+\frac{3}{8}x{4}
Simplifying the Right-Hand Side
Now that we have expanded the left-hand side of the equation, we can simplify the right-hand side. We can start by combining like terms and simplifying the expression.
Combining Like Terms
Combining like terms on the right-hand side of the equation, we get:
2xx=2x+2x^{2}
Equating the Two Expressions
Now that we have simplified both sides of the equation, we can equate the two expressions. This will give us a polynomial equation in terms of x.
Equating the Two Expressions
Equating the two expressions, we get:
1+\frac{3}{2}x{2}+\frac{3}{8}x{4}=2x+2x^{2}
Rearranging the Equation
To solve for x, we can rearrange the equation by subtracting 2x and 2x^{2} from both sides. This will give us a polynomial equation in terms of x.
Rearranging the Equation
Rearranging the equation, we get:
1+\frac{3}{2}x{2}+\frac{3}{8}x{4}-2x-2x^{2}=0
Combining Like Terms
Combining like terms on the left-hand side of the equation, we get:
1-\frac{1}{2}x{2}+\frac{3}{8}x{4}-2x=0
Factoring the Equation
To solve for x, we can try to factor the equation. This will give us a product of two binomials that can be set equal to zero.
Factoring the Equation
Factoring the equation, we get:
\left(1-\frac{1}{2}x{2}+{1}{4}x{4}\right)-2x=0
Factoring the First Term
Factoring the first term on the left-hand side of the equation, we get:
\left(1-\frac{1}{2}x{2}+\frac{1}{4}x{4}\right)=\left(1-\frac{1}{2}x{2}\right)\left(1+\frac{1}{2}x{2}\right)
Factoring the Second Term
Factoring the second term on the left-hand side of the equation, we get:
-2x=-2x
Setting the Two Expressions Equal to Zero
Now that we have factored the equation, we can set the two expressions equal to zero. This will give us a system of equations that we can solve for x.
Setting the Two Expressions Equal to Zero
Setting the two expressions equal to zero, we get:
\left(1-\frac{1}{2}x{2}+\frac{1}{4}x{4}\right)=0 -2x=0
Solving the First Equation
To solve the first equation, we can start by factoring the left-hand side. This will give us a product of two binomials that can be set equal to zero.
Solving the First Equation
Solving the first equation, we get:
\left(1-\frac{1}{2}x{2}+\frac{1}{4}x{4}\right)=\left(1-\frac{1}{2}x{2}\right)\left(1+\frac{1}{2}x{2}\right)=0
Setting Each Factor Equal to Zero
Now that we have factored the left-hand side of the equation, we can set each factor equal to zero. This will give us a system of equations that we can solve for x.
Setting Each Factor Equal to Zero
Setting each factor equal to zero, we get:
1-\frac{1}{2}x^{2}=0 1+\frac{1}{2}x^{2}=0
Solving the First Equation
To solve the first equation, we can start by adding \frac{1}{2}x^{2} to both sides. This will give us a quadratic equation in terms of x.
Solving the First Equation
Solving the first equation, we get:
1-\frac{1}{2}x{2}+\frac{1}{2}x{2}=1 1=1
Solving the Second Equation
To solve the second equation, we can start by subtracting 1 from both sides. This will give us a quadratic equation in terms of x.
Solving the Second Equation
Solving the second equation, we get:
1+\frac{1}{2}x{2}-1=\frac{1}{2}x{2} \frac{1}{2}x^{2}=0
Solving for x
Now that we have solved the two equations, we can solve for x. We can start by dividing both sides of the equation by \frac{1}{2}. This will give us a value for x.
Solving for x
Solving for x, we get:
\frac{1}{2}x^{2}=0 x^{2}=0 x=0
Conclusion
In this article, we have solved the equation \left(1+x{2}\right){1.5}=2+2xx using algebraic manipulation and factoring. We have shown that the solution to the equation is x=0. This demonstrates the power of algebraic manipulation and factoring in solving complex equations.
Final Answer
The final answer is x=0.
Introduction
In our previous article, we solved the equation \left(1+x{2}\right){1.5}=2+2xx using algebraic manipulation and factoring. In this article, we will answer some common questions that readers may have about the solution to this equation.
Q: What is the main concept behind solving this equation?
A: The main concept behind solving this equation is the use of algebraic manipulation and factoring to simplify the equation and isolate the variable x.
Q: Why did we use the binomial theorem to expand the left-hand side of the equation?
A: We used the binomial theorem to expand the left-hand side of the equation because it allowed us to simplify the expression and make it easier to work with.
Q: How did we factor the left-hand side of the equation?
A: We factored the left-hand side of the equation by recognizing that it was a product of two binomials. We then set each factor equal to zero and solved for x.
Q: What is the significance of the solution x=0?
A: The solution x=0 is significant because it represents the only value of x that satisfies the equation. This means that when x=0, the equation is true.
Q: Can we use this method to solve other equations?
A: Yes, we can use this method to solve other equations that involve exponents and variables. However, the specific steps and techniques used may vary depending on the equation.
Q: What are some common mistakes to avoid when solving equations like this?
A: Some common mistakes to avoid when solving equations like this include:
- Not expanding the left-hand side of the equation properly
- Not factoring the left-hand side of the equation correctly
- Not setting each factor equal to zero and solving for x
- Not checking the solution to make sure it satisfies the original equation
Q: How can we apply this method to real-world problems?
A: This method can be applied to real-world problems that involve solving equations with exponents and variables. For example, in physics, we may need to solve equations that involve the motion of objects or the behavior of electrical circuits.
Q: What are some other techniques we can use to solve equations like this?
A: Some other techniques we can use to solve equations like this include:
- Using the quadratic formula to solve quadratic equations
- Using the rational root theorem to find possible rational roots of a polynomial equation
- Using synthetic division to divide polynomials and find roots
Q: Can we use technology to solve equations like this?
A: Yes, we can use technology such as calculators or computer software to solve equations like this. However, it's always a good idea to understand the underlying math and be able to solve the equation by hand.
Conclusion
In this article, we have answered some common questions about solving the equation \left(1+x{2}\right){1.5}=2+2xx. We have discussed the main concept behind solving this equation, the significance of the solution x=0, and some common mistakes to avoid. We have also discussed how to apply this method to real-world problems and some other techniques we can use to solve equations like this.
Final Answer
The final answer is x=0.