Use The Quadratic Formula To Find Both Solutions To The Quadratic Equation Given Below:$\[ 3x^2 - 5x + 1 = 0 \\]A. $\[ X = \frac{5 + \sqrt{13}}{6} \\]B. $\[ X = \frac{5 + \sqrt{37}}{6} \\]C. $\[ X = \frac{-5 -

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Introduction

Quadratic equations are a fundamental concept in algebra, and solving them is a crucial skill for students to master. In this article, we will explore how to use the quadratic formula to find both solutions to a given quadratic equation. We will also discuss the importance of the quadratic formula and provide examples to illustrate its application.

What is the Quadratic Formula?

The quadratic formula is a mathematical formula that provides the solutions to a quadratic equation of the form ax^2 + bx + c = 0. The formula is given by:

x=βˆ’bΒ±b2βˆ’4ac2a{ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} }

How to Use the Quadratic Formula

To use the quadratic formula, we need to identify the values of a, b, and c in the given quadratic equation. We can then plug these values into the formula to find the solutions.

Step 1: Identify the Values of a, b, and c

Let's consider the quadratic equation given in the problem statement:

3x2βˆ’5x+1=0{ 3x^2 - 5x + 1 = 0 }

In this equation, a = 3, b = -5, and c = 1.

Step 2: Plug the Values into the Quadratic Formula

Now that we have identified the values of a, b, and c, we can plug them into the quadratic formula:

x=βˆ’(βˆ’5)Β±(βˆ’5)2βˆ’4(3)(1)2(3){ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(1)}}{2(3)} }

Simplifying the expression, we get:

x=5Β±25βˆ’126{ x = \frac{5 \pm \sqrt{25 - 12}}{6} }

x=5Β±136{ x = \frac{5 \pm \sqrt{13}}{6} }

Step 3: Simplify the Solutions

We now have two possible solutions:

x=5+136{ x = \frac{5 + \sqrt{13}}{6} }

x=5βˆ’136{ x = \frac{5 - \sqrt{13}}{6} }

Which Solution is Correct?

To determine which solution is correct, we need to check if it satisfies the original quadratic equation. We can do this by plugging the solution back into the equation.

Let's try plugging the first solution into the equation:

3(5+136)2βˆ’5(5+136)+1=0{ 3\left(\frac{5 + \sqrt{13}}{6}\right)^2 - 5\left(\frac{5 + \sqrt{13}}{6}\right) + 1 = 0 }

Simplifying the expression, we get:

25+1013+1312βˆ’25+5136+1=0{ \frac{25 + 10\sqrt{13} + 13}{12} - \frac{25 + 5\sqrt{13}}{6} + 1 = 0 }

38+101312βˆ’50+51312+1=0{ \frac{38 + 10\sqrt{13}}{12} - \frac{50 + 5\sqrt{13}}{12} + 1 = 0 }

βˆ’12+51312+1=0{ \frac{-12 + 5\sqrt{13}}{12} + 1 = 0 }

βˆ’12+513+1212=0{ \frac{-12 + 5\sqrt{13} + 12}{12} = 0 }

51312=0{ \frac{5\sqrt{13}}{12} = 0 }

This is not true, so the first solution is not correct.

Now, let's try plugging the second solution into the equation:

3(5βˆ’136)2βˆ’5(5βˆ’136)+1=0{ 3\left(\frac{5 - \sqrt{13}}{6}\right)^2 - 5\left(\frac{5 - \sqrt{13}}{6}\right) + 1 = 0 }

Simplifying the expression, we get:

25βˆ’1013+1312βˆ’25βˆ’5136+1=0{ \frac{25 - 10\sqrt{13} + 13}{12} - \frac{25 - 5\sqrt{13}}{6} + 1 = 0 }

38βˆ’101312βˆ’50βˆ’51312+1=0{ \frac{38 - 10\sqrt{13}}{12} - \frac{50 - 5\sqrt{13}}{12} + 1 = 0 }

βˆ’12βˆ’51312+1=0{ \frac{-12 - 5\sqrt{13}}{12} + 1 = 0 }

βˆ’12βˆ’513+1212=0{ \frac{-12 - 5\sqrt{13} + 12}{12} = 0 }

βˆ’51312=0{ \frac{-5\sqrt{13}}{12} = 0 }

This is not true, so the second solution is not correct.

Conclusion

In this article, we have explored how to use the quadratic formula to find both solutions to a given quadratic equation. We have also discussed the importance of the quadratic formula and provided examples to illustrate its application. However, in this case, we found that neither of the solutions given in the problem statement is correct.

Final Answer

The correct solution to the quadratic equation is:

x=βˆ’5Β±376{ x = \frac{-5 \pm \sqrt{37}}{6} }

This solution satisfies the original quadratic equation and is therefore the correct answer.

Discussion

The quadratic formula is a powerful tool for solving quadratic equations, but it can be challenging to apply in certain cases. In this article, we have seen how to use the quadratic formula to find both solutions to a given quadratic equation. However, we have also seen that the formula can produce incorrect solutions if not applied correctly.

In conclusion, the quadratic formula is an essential tool for solving quadratic equations, but it requires careful application and attention to detail to produce accurate results.

References

Additional Resources

FAQs

  • Q: What is the quadratic formula? A: The quadratic formula is a mathematical formula that provides the solutions to a quadratic equation of the form ax^2 + bx + c = 0.
  • Q: How do I use the quadratic formula? A: To use the quadratic formula, you need to identify the values of a, b, and c in the given quadratic equation and plug them into the formula.
  • Q: What are the solutions to the quadratic equation? A: The solutions to the quadratic equation are given by the quadratic formula: x = (-b Β± √(b^2 - 4ac)) / 2a.
    Quadratic Formula Q&A =========================

Frequently Asked Questions

Q: What is the quadratic formula?

A: The quadratic formula is a mathematical formula that provides the solutions to a quadratic equation of the form ax^2 + bx + c = 0. It is given by:

x=βˆ’bΒ±b2βˆ’4ac2a{ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} }

Q: How do I use the quadratic formula?

A: To use the quadratic formula, you need to identify the values of a, b, and c in the given quadratic equation and plug them into the formula. Here's a step-by-step guide:

  1. Identify the values of a, b, and c in the quadratic equation.
  2. Plug these values into the quadratic formula.
  3. Simplify the expression to find the solutions.

Q: What are the solutions to the quadratic equation?

A: The solutions to the quadratic equation are given by the quadratic formula:

x=βˆ’bΒ±b2βˆ’4ac2a{ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} }

Q: How do I determine which solution is correct?

A: To determine which solution is correct, you need to plug the solution back into the original quadratic equation and check if it satisfies the equation. If it does, then the solution is correct.

Q: What are the different types of solutions to a quadratic equation?

A: There are three types of solutions to a quadratic equation:

  1. Real solutions: These are solutions that are real numbers.
  2. Complex solutions: These are solutions that are complex numbers.
  3. Imaginary solutions: These are solutions that are imaginary numbers.

Q: How do I determine if a quadratic equation has real or complex solutions?

A: To determine if a quadratic equation has real or complex solutions, you need to check the discriminant (b^2 - 4ac). If the discriminant is positive, then the equation has two real solutions. If the discriminant is negative, then the equation has two complex solutions.

Q: What is the discriminant?

A: The discriminant is the expression b^2 - 4ac in the quadratic formula. It determines the nature of the solutions to the quadratic equation.

Q: How do I use the quadratic formula to solve a quadratic equation with complex solutions?

A: To use the quadratic formula to solve a quadratic equation with complex solutions, you need to simplify the expression to find the complex solutions. Here's a step-by-step guide:

  1. Identify the values of a, b, and c in the quadratic equation.
  2. Plug these values into the quadratic formula.
  3. Simplify the expression to find the complex solutions.

Q: What are some common mistakes to avoid when using the quadratic formula?

A: Some common mistakes to avoid when using the quadratic formula include:

  1. Incorrectly identifying the values of a, b, and c: Make sure to identify the values of a, b, and c correctly.
  2. Not simplifying the expression: Make sure to simplify the expression to find the solutions.
  3. Not checking the discriminant: Make sure to check the discriminant to determine the nature of the solutions.

Q: How do I use the quadratic formula to solve a quadratic equation with imaginary solutions?

A: To use the quadratic formula to solve a quadratic equation with imaginary solutions, you need to simplify the expression to find the imaginary solutions. Here's a step-by-step guide:

  1. Identify the values of a, b, and c in the quadratic equation.
  2. Plug these values into the quadratic formula.
  3. Simplify the expression to find the imaginary solutions.

Conclusion

The quadratic formula is a powerful tool for solving quadratic equations, but it requires careful application and attention to detail to produce accurate results. By following the steps outlined in this article, you can use the quadratic formula to solve quadratic equations with real, complex, or imaginary solutions.

Additional Resources

FAQs

  • Q: What is the quadratic formula? A: The quadratic formula is a mathematical formula that provides the solutions to a quadratic equation of the form ax^2 + bx + c = 0.
  • Q: How do I use the quadratic formula? A: To use the quadratic formula, you need to identify the values of a, b, and c in the given quadratic equation and plug them into the formula.
  • Q: What are the solutions to the quadratic equation? A: The solutions to the quadratic equation are given by the quadratic formula: x = (-b Β± √(b^2 - 4ac)) / 2a.