Solve For P P P And Q Q Q In The Equation: P 2 + 2 P Q + Q 2 = 1 P^2 + 2pq + Q^2 = 1 P 2 + 2 Pq + Q 2 = 1 Choose The Correct Value:A. 0.49 B. 0.30 C. 0.55 D. 0.70

Introduction

Quadratic equations are a fundamental concept in mathematics, and solving them is a crucial skill for students and professionals alike. In this article, we will focus on solving a specific quadratic equation, $p^2 + 2pq + q^2 = 1$, and find the correct values of $p$ and $q$. We will break down the solution into manageable steps, making it easy to understand and follow.

Understanding the Equation

The given equation is a quadratic equation in the form of $ax^2 + bx + c = 0$, where $a = 1$, $b = 2p$, and $c = q^2$. To solve for $p$ and $q$, we need to manipulate the equation to isolate the variables.

Step 1: Factor the Equation

The equation $p^2 + 2pq + q^2 = 1$ can be factored as $(p + q)^2 = 1$. This is because the left-hand side of the equation is a perfect square trinomial, which can be factored into the square of a binomial.

Step 2: Take the Square Root

Taking the square root of both sides of the equation, we get $p + q = \pm 1$. This is because the square root of a number can be either positive or negative.

Step 3: Solve for $p$ and $q$

Now that we have the equation $p + q = \pm 1$, we can solve for $p$ and $q$ by using the fact that $p^2 + 2pq + q^2 = 1$. We can rewrite this equation as $(p + q)^2 = 1$, which is equal to $p^2 + 2pq + q^2 = 1$.

Case 1: $p + q = 1$

If $p + q = 1$, then we can substitute this value into the equation $p^2 + 2pq + q^2 = 1$. This gives us $p^2 + 2p(1) + 1^2 = 1$, which simplifies to $p^2 + 2p + 1 = 1$. Subtracting 1 from both sides, we get $p^2 + 2p = 0$. Factoring out $p$, we get $p(p + 2) = 0$. This gives us two possible solutions: $p = 0$ or $p + 2 = 0$, which implies $p = -2$.

Case 2: $p + q = -1$

If $p + q = -1$, then we can substitute this value into the equation $p^2 + 2pq + q^2 = 1$. This gives us $p^2 + 2p(-1) + (-1)^2 = 1$, which simplifies to $p^2 - 2p + 1 = 1$. Subtracting 1 from both sides, we get $p^2 - 2p = 0$. Factoring out $p$, we get $p(p - 2) = 0$. This gives us two possible solutions: $p = 0$ or $p - 2 = 0$, which implies $p = 2$.

Conclusion

In conclusion, we have solved the quadratic equation $p^2 + 2pq + q^2 = 1$ and found the correct values of $p$ and $q$. The solutions are $p = 0$ and $q = 1$, or $p = -2$ and $q = 1$, or $p = 0$ and $q = -1$, or $p = 2$ and $q = -1$. However, we need to choose the correct value from the given options.

Choosing the Correct Value

Looking at the options, we can see that the correct value is $p = 0.49$ and $q = 0.70$. This is because the equation $p^2 + 2pq + q^2 = 1$ can be rewritten as $(p + q)^2 = 1$, which implies $p + q = \pm 1$. Substituting $p = 0.49$ and $q = 0.70$ into the equation, we get $(0.49 + 0.70)^2 = 1$, which simplifies to $1.19^2 = 1$. This is a true statement, so the correct value is $p = 0.49$ and $q = 0.70$.

Final Answer

$$

Introduction

Quadratic equations are a fundamental concept in mathematics, and solving them is a crucial skill for students and professionals alike. In this article, we will provide a Q&A guide to help you understand and solve quadratic equations.

Q: What is a quadratic equation?

A: A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable is two. It is typically written in the form of $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants.

Q: How do I solve a quadratic equation?

A: To solve a quadratic equation, you can use the following methods:

• Factoring: If the equation can be factored into the product of two binomials, you can solve it by setting each binomial equal to zero.
• Quadratic formula: If the equation cannot be factored, you can use the quadratic formula to solve it. The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Graphing: You can also solve a quadratic equation by graphing the related function and finding the x-intercepts.

Q: What is the quadratic formula?

A: The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. This formula can be used to solve any quadratic equation in the form of $ax^2 + bx + c = 0$.

Q: How do I use the quadratic formula?

A: To use the quadratic formula, you need to plug in the values of $a$, $b$, and $c$ into the formula. Then, simplify the expression and solve for $x$.

Q: What is the difference between the quadratic formula and factoring?

A: The quadratic formula and factoring are two different methods for solving quadratic equations. Factoring involves finding the factors of the quadratic expression, while the quadratic formula involves using a formula to solve the equation.

Q: Can I use the quadratic formula to solve any quadratic equation?

A: Yes, the quadratic formula can be used to solve any quadratic equation in the form of $ax^2 + bx + c = 0$. However, if the equation can be factored, it is often easier to use factoring to solve it.

Q: How do I choose between factoring and the quadratic formula?

A: To choose between factoring and the quadratic formula, you need to consider the complexity of the equation. If the equation can be factored easily, it is often better to use factoring. However, if the equation is complex or cannot be factored, the quadratic formula may be a better option.

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

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

• Not checking the solutions for extraneous solutions
• Not simplifying the expression before solving for $x$
• Not using the correct formula or method for the given equation

Conclusion

In conclusion, quadratic equations are a fundamental concept in mathematics, and solving them is a crucial skill for students and professionals alike. By understanding the different methods for solving quadratic equations, including factoring and the quadratic formula, you can solve a wide range of equations and problems.

Final Tips

• Practice, practice, practice: The more you practice solving quadratic equations, the more comfortable you will become with the different methods and formulas.
• Use the correct formula or method for the given equation: Make sure to use the correct formula or method for the given equation to avoid mistakes.
• Check your solutions: Always check your solutions for extraneous solutions to ensure that they are valid.

Common Quadratic Equations

Here are some common quadratic equations that you may encounter:

• $x^2 + 5x + 6 = 0$
• $x^2 - 7x + 12 = 0$
• $x^2 + 2x - 15 = 0$

Solving Quadratic Equations with the Quadratic Formula

Here are some examples of solving quadratic equations using the quadratic formula:

• $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
• $x = \frac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)}$
• $x = \frac{-5 \pm \sqrt{25 - 24}}{2}$
• $x = \frac{-5 \pm \sqrt{1}}{2}$
• $x = \frac{-5 \pm 1}{2}$

Solving Quadratic Equations by Factoring

Here are some examples of solving quadratic equations by factoring:

• $x^2 + 5x + 6 = (x + 2)(x + 3) = 0$
• $x = -2$ or $x = -3$
• $x^2 - 7x + 12 = (x - 3)(x - 4) = 0$
• $x = 3$ or $x = 4$

Conclusion

In conclusion, quadratic equations are a fundamental concept in mathematics, and solving them is a crucial skill for students and professionals alike. By understanding the different methods for solving quadratic equations, including factoring and the quadratic formula, you can solve a wide range of equations and problems.