Select The Two Values Of $x$ That Are Roots Of This Equation.$2x^2 + 11x + 15 = 0$A. $ X = − 6 X = -6 X = − 6 [/tex] B. $x = -3$ C. $x = -5$ D. $x = -\frac{5}{2}$

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Introduction

Quadratic equations are a fundamental concept in mathematics, and solving them is a crucial skill for students to master. In this article, we will focus on solving a specific quadratic equation, $2x^2 + 11x + 15 = 0$, and find the two values of $x$ that are roots of this equation.

Understanding Quadratic Equations

A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable is two. The general form of a quadratic equation is $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants, and $x$ is the variable. In our given equation, $a = 2$, $b = 11$, and $c = 15$.

The Quadratic Formula

To solve a quadratic equation, we can use the quadratic formula, which is given by:

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

This formula will give us two solutions for $x$, which are the roots of the equation.

Applying the Quadratic Formula

Now, let's apply the quadratic formula to our given equation, $2x^2 + 11x + 15 = 0$.

First, we identify the values of $a$, $b$, and $c$, which are $a = 2$, $b = 11$, and $c = 15$.

Next, we plug these values into the quadratic formula:

x=−11±112−4(2)(15)2(2)x = \frac{-11 \pm \sqrt{11^2 - 4(2)(15)}}{2(2)}

Simplifying the expression under the square root, we get:

x=−11±121−1204x = \frac{-11 \pm \sqrt{121 - 120}}{4}

x=−11±14x = \frac{-11 \pm \sqrt{1}}{4}

x=−11±14x = \frac{-11 \pm 1}{4}

Finding the Roots

Now, we have two possible solutions for $x$, which are:

x=−11+14=−104=−52x = \frac{-11 + 1}{4} = \frac{-10}{4} = -\frac{5}{2}

x=−11−14=−124=−3x = \frac{-11 - 1}{4} = \frac{-12}{4} = -3

Conclusion

In this article, we solved a quadratic equation, $2x^2 + 11x + 15 = 0$, using the quadratic formula. We found two values of $x$ that are roots of this equation, which are $x = -\frac{5}{2}$ and $x = -3$. These values are the solutions to the equation, and they satisfy the equation when substituted back into the original equation.

Answer

Based on our calculations, the correct answers are:

  • A. $x = -6$ is not a solution to the equation.
  • B. $x = -3$ is a solution to the equation.
  • C. $x = -5$ is not a solution to the equation.
  • D. $x = -\frac{5}{2}$ is a solution to the equation.

Therefore, the correct answers are B and D.

Introduction

Quadratic equations are a fundamental concept in mathematics, and solving them is a crucial skill for students to master. In our previous article, we solved a quadratic equation, $2x^2 + 11x + 15 = 0$, using the quadratic formula. In this article, we will provide a comprehensive Q&A guide to help students understand quadratic equations and how to solve them.

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. The general form of a quadratic equation is $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants, and $x$ is the variable.

Q: What is the quadratic formula?

A: The quadratic formula is a mathematical formula that gives us the solutions to a quadratic equation. It is given by:

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

Q: How do I apply the quadratic formula?

A: To apply the quadratic formula, you need to identify the values of $a$, $b$, and $c$ in the quadratic equation. Then, you plug these values into the quadratic formula and simplify the expression.

Q: What is the difference between the two solutions given by the quadratic formula?

A: The two solutions given by the quadratic formula are:

x=−b+b2−4ac2ax = \frac{-b + \sqrt{b^2 - 4ac}}{2a}

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

The difference between these two solutions is the sign of the square root term. If the square root term is positive, the solutions are real and distinct. If the square root term is zero, the solutions are real and equal. If the square root term is negative, the solutions are complex.

Q: How do I determine if the solutions are real or complex?

A: To determine if the solutions are real or complex, you need to check the sign of the square root term. If the square root term is positive, the solutions are real and distinct. If the square root term is zero, the solutions are real and equal. If the square root term is negative, the solutions are complex.

Q: What is the significance of the discriminant in the quadratic formula?

A: The discriminant is the expression under the square root in the quadratic formula, which is $b^2 - 4ac$. The discriminant determines the nature of the solutions. If the discriminant is positive, the solutions are real and distinct. If the discriminant is zero, the solutions are real and equal. If the discriminant is negative, the solutions are complex.

Q: How do I simplify the expression under the square root in the quadratic formula?

A: To simplify the expression under the square root, you need to factor the expression and simplify it. You can also use the difference of squares formula to simplify the expression.

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

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

  • Not identifying the values of $a$, $b$, and $c$ correctly
  • Not simplifying the expression under the square root correctly
  • Not checking the sign of the square root term
  • Not determining the nature of the solutions correctly

Conclusion

In this article, we provided a comprehensive Q&A guide to help students understand quadratic equations and how to solve them. We covered topics such as the quadratic formula, applying the quadratic formula, and determining the nature of the solutions. We also discussed common mistakes to avoid when solving quadratic equations. By following this guide, students should be able to solve quadratic equations with confidence.