In Each Of The Following Equations, Solve For $x$ In Terms Of $p$:1. $4 X^2 = 16 P^2$2. $4 X^2 = 16 P X$3. $x^2 - 5 P X + 6 P^2 = 0$4. $\left(\frac{x}{p}\right)^2 = 9 P^2$

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Solving Equations for x in Terms of p

In mathematics, solving equations for a specific variable is a fundamental concept that helps us understand the relationship between different variables. In this article, we will focus on solving equations for x in terms of p, which is a common problem in algebra and calculus. We will go through four different equations and solve for x in terms of p.

Equation 1: 4x2=16p24 x^2 = 16 p^2

To solve for x in terms of p, we need to isolate x on one side of the equation. We can start by dividing both sides of the equation by 4, which gives us:

x2=4p2x^2 = 4 p^2

Next, we can take the square root of both sides of the equation, which gives us:

x=±2px = \pm 2 p

Therefore, the solution to the equation 4x2=16p24 x^2 = 16 p^2 is x=±2px = \pm 2 p.

Equation 2: 4x2=16px4 x^2 = 16 p x

To solve for x in terms of p, we need to isolate x on one side of the equation. We can start by dividing both sides of the equation by 4, which gives us:

x2=4pxx^2 = 4 p x

Next, we can divide both sides of the equation by x, which gives us:

x=4px = 4 p

However, we need to check if this solution is valid. If x = 0, then the left-hand side of the equation is 0, but the right-hand side is not. Therefore, x cannot be 0, and we can conclude that the solution to the equation 4x2=16px4 x^2 = 16 p x is x=4px = 4 p.

Equation 3: x2−5px+6p2=0x^2 - 5 p x + 6 p^2 = 0

To solve for x in terms of p, we can use the quadratic formula. The quadratic formula is given by:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a}

In this case, a = 1, b = -5p, and c = 6p^2. Plugging these values into the quadratic formula, we get:

x=5p±(−5p)2−4(1)(6p2)2(1)x = \frac{5 p \pm \sqrt{(-5 p)^2 - 4 (1) (6 p^2)}}{2 (1)}

Simplifying the expression under the square root, we get:

x=5p±25p2−24p22x = \frac{5 p \pm \sqrt{25 p^2 - 24 p^2}}{2}

x=5p±p22x = \frac{5 p \pm \sqrt{p^2}}{2}

x=5p±p2x = \frac{5 p \pm p}{2}

Therefore, the solutions to the equation x2−5px+6p2=0x^2 - 5 p x + 6 p^2 = 0 are x=5p+p2x = \frac{5 p + p}{2} and x=5p−p2x = \frac{5 p - p}{2}.

Equation 4: (xp)2=9p2\left(\frac{x}{p}\right)^2 = 9 p^2

To solve for x in terms of p, we can start by taking the square root of both sides of the equation, which gives us:

xp=±3p\frac{x}{p} = \pm 3 p

Next, we can multiply both sides of the equation by p, which gives us:

x=±3p2x = \pm 3 p^2

Therefore, the solution to the equation (xp)2=9p2\left(\frac{x}{p}\right)^2 = 9 p^2 is x=±3p2x = \pm 3 p^2.

Conclusion

In this article, we have solved four different equations for x in terms of p. We have used various techniques, including factoring, the quadratic formula, and algebraic manipulation, to isolate x on one side of the equation. The solutions to the equations are:

  • x=±2px = \pm 2 p for the equation 4x2=16p24 x^2 = 16 p^2
  • x=4px = 4 p for the equation 4x2=16px4 x^2 = 16 p x
  • x=5p+p2x = \frac{5 p + p}{2} and x=5p−p2x = \frac{5 p - p}{2} for the equation x2−5px+6p2=0x^2 - 5 p x + 6 p^2 = 0
  • x=±3p2x = \pm 3 p^2 for the equation (xp)2=9p2\left(\frac{x}{p}\right)^2 = 9 p^2

These solutions demonstrate the importance of algebraic manipulation and the quadratic formula in solving equations for x in terms of p.
Frequently Asked Questions: Solving Equations for x in Terms of p

In the previous article, we solved four different equations for x in terms of p. However, we know that there are many more questions and doubts that readers may have. In this article, we will address some of the most frequently asked questions about solving equations for x in terms of p.

Q: What is the difference between solving for x in terms of p and solving for x in terms of other variables?

A: Solving for x in terms of p means that we are expressing x as a function of p, where p is a given variable. This is different from solving for x in terms of other variables, such as y or z, where x is a function of those variables.

Q: How do I know which method to use when solving for x in terms of p?

A: The method you use will depend on the type of equation you are solving. For example, if you have a quadratic equation, you may want to use the quadratic formula. If you have a linear equation, you may want to use algebraic manipulation.

Q: What is the quadratic formula, and how do I use it?

A: The quadratic formula is a mathematical formula that is used to solve quadratic equations. It is given by:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a}

To use the quadratic formula, you need to identify the values of a, b, and c in the equation. Then, you can plug these values into the formula and simplify.

Q: How do I handle complex solutions when solving for x in terms of p?

A: Complex solutions occur when the square root of a negative number is involved. To handle complex solutions, you need to use the imaginary unit, i, which is defined as the square root of -1. For example, if you have a solution of the form x = a + bi, where a and b are real numbers, then you can write the solution as x = a + bi.

Q: Can I use technology, such as calculators or computer software, to solve equations for x in terms of p?

A: Yes, you can use technology to solve equations for x in terms of p. Many calculators and computer software programs, such as graphing calculators and computer algebra systems, can solve equations and provide solutions.

Q: How do I check my solutions to make sure they are correct?

A: To check your solutions, you need to plug the solutions back into the original equation and simplify. If the solutions satisfy the equation, then they are correct. If the solutions do not satisfy the equation, then you need to recheck your work and try again.

Q: What are some common mistakes to avoid when solving for x in terms of p?

A: Some common mistakes to avoid when solving for x in terms of p include:

  • Not following the order of operations
  • Not simplifying expressions
  • Not checking solutions
  • Not using the correct method for the type of equation

Q: How do I apply what I have learned to real-world problems?

A: To apply what you have learned to real-world problems, you need to be able to identify the type of equation and the variables involved. Then, you can use the methods and techniques you have learned to solve the equation and find the solution.

Q: What are some additional resources that I can use to learn more about solving equations for x in terms of p?

A: Some additional resources that you can use to learn more about solving equations for x in terms of p include:

  • Textbooks and online resources
  • Video tutorials and online courses
  • Practice problems and worksheets
  • Real-world applications and examples

By following these tips and resources, you can improve your skills and knowledge in solving equations for x in terms of p.