Solve For X X X In The Equation X 2 + 11 X + 121 4 = 125 4 X^2 + 11x + \frac{121}{4} = \frac{125}{4} X 2 + 11 X + 4 121 ​ = 4 125 ​ .A. X = − 11 ± 25 2 X = -11 \pm \frac{25}{2} X = − 11 ± 2 25 ​ B. X = − 11 2 ± 25 2 X = -\frac{11}{2} \pm \frac{25}{2} X = − 2 11 ​ ± 2 25 ​ C. X = − 11 ± 5 5 2 X = -11 \pm \frac{5 \sqrt{5}}{2} X = − 11 ± 2 5 5 ​ ​ D. $x = -\frac{11}{2}

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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, x2+11x+1214=1254x^2 + 11x + \frac{121}{4} = \frac{125}{4}, and explore the different methods and techniques used to find the solutions.

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 ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are constants. In our given equation, x2+11x+1214=1254x^2 + 11x + \frac{121}{4} = \frac{125}{4}, we can rewrite it in the standard form as x2+11x+1214−1254=0x^2 + 11x + \frac{121}{4} - \frac{125}{4} = 0.

Solving the Quadratic Equation

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

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

In our equation, a=1a = 1, b=11b = 11, and c=1214−1254=−44=−1c = \frac{121}{4} - \frac{125}{4} = -\frac{4}{4} = -1. Plugging these values into the quadratic formula, we get:

x=−11±112−4(1)(−1)2(1)x = \frac{-11 \pm \sqrt{11^2 - 4(1)(-1)}}{2(1)}

x=−11±121+42x = \frac{-11 \pm \sqrt{121 + 4}}{2}

x=−11±1252x = \frac{-11 \pm \sqrt{125}}{2}

x=−11±552x = \frac{-11 \pm 5\sqrt{5}}{2}

Analyzing the Solutions

Now that we have the solutions, let's analyze them. We have two possible values for xx, which are x=−11+552x = -11 + \frac{5\sqrt{5}}{2} and x=−11−552x = -11 - \frac{5\sqrt{5}}{2}. These values represent the two solutions to the quadratic equation.

Conclusion

In this article, we solved a quadratic equation using the quadratic formula. We started by rewriting the equation in the standard form, then applied the quadratic formula to find the solutions. The solutions were analyzed, and we found that there are two possible values for xx. This demonstrates the importance of quadratic equations in mathematics and their applications in various fields.

Final Answer

The final answer to the equation x2+11x+1214=1254x^2 + 11x + \frac{121}{4} = \frac{125}{4} is:

x=−11±552x = -11 \pm \frac{5\sqrt{5}}{2}

This is the correct solution to the equation, and it matches option C.

Discussion

The discussion category for this article is mathematics, specifically quadratic equations. The article provides a step-by-step guide on solving quadratic equations using the quadratic formula. The solutions are analyzed, and the final answer is provided.

Recommendations

For students and professionals who want to improve their skills in solving quadratic equations, we recommend the following:

  • Practice solving quadratic equations using the quadratic formula.
  • Analyze the solutions to understand the concept of quadratic equations.
  • Apply quadratic equations to real-world problems to see their practical applications.

By following these recommendations, you can improve your skills in solving quadratic equations and apply them to various fields.

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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 answer some frequently asked questions about quadratic equations, providing a comprehensive guide to help you understand and solve these equations.

Q1: What is a quadratic equation?

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 ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are constants.

Q2: How do I solve a quadratic equation?

There are several methods to solve a quadratic equation, including:

  • Factoring: If the quadratic expression can be factored into the product of two binomials, we can solve for the variable.
  • Quadratic formula: The quadratic formula is given by x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where aa, bb, and cc are the coefficients of the quadratic equation.
  • Graphing: We can graph the quadratic function and find the x-intercepts, which represent the solutions to the equation.

Q3: What is the quadratic formula?

The quadratic formula is given by x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where aa, bb, and cc are the coefficients of the quadratic equation.

Q4: How do I apply the quadratic formula?

To apply the quadratic formula, we need to identify the values of aa, bb, and cc in the quadratic equation. Then, we plug these values into the quadratic formula and simplify to find the solutions.

Q5: What are the solutions to a quadratic equation?

The solutions to a quadratic equation are the values of the variable that satisfy the equation. These solutions can be real or complex numbers.

Q6: How do I determine the number of solutions to a quadratic equation?

The number of solutions to a quadratic equation depends on the discriminant, which is given by b2−4acb^2 - 4ac. If the discriminant is positive, the equation has two distinct solutions. If the discriminant is zero, the equation has one repeated solution. If the discriminant is negative, the equation has no real solutions.

Q7: What is the difference between a quadratic equation and a linear equation?

A quadratic equation is a polynomial equation of degree two, while a linear equation is a polynomial equation of degree one. The general form of a linear equation is ax+b=0ax + b = 0, where aa and bb are constants.

Q8: Can I use the quadratic formula to solve a linear equation?

No, the quadratic formula is used to solve quadratic equations, not linear equations. To solve a linear equation, we can use the formula x=−bax = -\frac{b}{a}.

Q9: How do I apply quadratic equations to real-world problems?

Quadratic equations have numerous applications in various fields, including physics, engineering, economics, and computer science. Some examples of real-world problems that involve quadratic equations include:

  • Projectile motion: The trajectory of a projectile under the influence of gravity can be modeled using a quadratic equation.
  • Optimization: Quadratic equations can be used to optimize functions and find the maximum or minimum value.
  • Data analysis: Quadratic equations can be used to model and analyze data in various fields.

Q10: What are some common mistakes to avoid when solving quadratic equations?

Some common mistakes to avoid when solving quadratic equations include:

  • Not simplifying the quadratic expression before applying the quadratic formula.
  • Not checking the discriminant to determine the number of solutions.
  • Not using the correct formula to solve the equation.

Conclusion

In this article, we answered some frequently asked questions about quadratic equations, providing a comprehensive guide to help you understand and solve these equations. We covered topics such as the quadratic formula, solutions, and applications of quadratic equations. By following the tips and recommendations in this article, you can improve your skills in solving quadratic equations and apply them to various fields.