Solve $3x^2 + 4x = 2$A. $\frac{2 \pm \sqrt{10}}{6}$B. $\frac{-2 \pm 2\sqrt{10}}{3}$C. $\frac{-2 \pm \sqrt{10}}{3}$D. $\frac{-4 \pm \sqrt{10}}{3}$

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Introduction

Solving quadratic equations is a fundamental concept in mathematics, and it is essential to understand the various methods of solving them. In this article, we will focus on solving the quadratic equation $3x^2 + 4x = 2$ using the quadratic formula. The quadratic formula is a powerful tool for solving quadratic equations, and it is widely used in various fields such as physics, engineering, and economics.

Understanding the Quadratic Formula

The quadratic formula is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a$, $b$, and $c$ are the coefficients of the quadratic equation. In this case, we have:

$a = 3$, $b = 4$, and $c = -2$

Applying the Quadratic Formula

Substituting the values of $a$, $b$, and $c$ into the quadratic formula, we get:

$x = \frac{-4 \pm \sqrt{4^2 - 4(3)(-2)}}{2(3)}$

$x = \frac{-4 \pm \sqrt{16 + 24}}{6}$

$x = \frac{-4 \pm \sqrt{40}}{6}$

$x = \frac{-4 \pm 2\sqrt{10}}{6}$

Simplifying the Solution

We can simplify the solution by dividing both the numerator and the denominator by 2:

$x = \frac{-2 \pm \sqrt{10}}{3}$

Conclusion

In this article, we have solved the quadratic equation $3x^2 + 4x = 2$ using the quadratic formula. We have applied the formula and simplified the solution to get the final answer. The correct solution is:

$x = \frac{-2 \pm \sqrt{10}}{3}$

This solution is in the form of a quadratic formula, which is a powerful tool for solving quadratic equations.

Discussion

The quadratic formula is a fundamental concept in mathematics, and it is widely used in various fields such as physics, engineering, and economics. The formula is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a$, $b$, and $c$ are the coefficients of the quadratic equation. In this case, we have:

$a = 3$, $b = 4$, and $c = -2$

The quadratic formula is a powerful tool for solving quadratic equations, and it is widely used in various fields such as physics, engineering, and economics.

Example

Let's consider another example of a quadratic equation:

$x^2 + 5x + 6 = 0$

We can solve this equation using the quadratic formula:

$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)}$

$x = \frac{-5 \pm \sqrt{25 - 24}}{2}$

$x = \frac{-5 \pm \sqrt{1}}{2}$

$x = \frac{-5 \pm 1}{2}$

$x = \frac{-4}{2}$ or $x = \frac{-6}{2}$

$x = -2$ or $x = -3$

Applications

The quadratic formula has numerous applications in various fields such as physics, engineering, and economics. Some of the applications of the quadratic formula include:

• Physics: The quadratic formula is used to solve problems involving motion, such as the trajectory of a projectile or the motion of an object under the influence of gravity.
• Engineering: The quadratic formula is used to solve problems involving the design of structures, such as bridges or buildings.
• Economics: The quadratic formula is used to solve problems involving the behavior of economic systems, such as the behavior of supply and demand.

Conclusion

In this article, we have solved the quadratic equation $3x^2 + 4x = 2$ using the quadratic formula. We have applied the formula and simplified the solution to get the final answer. The correct solution is:

$x = \frac{-2 \pm \sqrt{10}}{3}$

This solution is in the form of a quadratic formula, which is a powerful tool for solving quadratic equations. The quadratic formula has numerous applications in various fields such as physics, engineering, and economics.

References

• "Algebra" by Michael Artin
• "Calculus" by Michael Spivak
• "Linear Algebra and Its Applications" by Gilbert Strang

Further Reading

• "Quadratic Equations" by Math Open Reference
• "Quadratic Formula" by Wolfram MathWorld
• "Solving Quadratic Equations" by Khan Academy
  

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Introduction

In our previous article, we solved the quadratic equation $3x^2 + 4x = 2$ using the quadratic formula. In this article, we will answer some of the frequently asked questions related to solving quadratic equations.

Q&A

Q1: What is the quadratic formula?

A1: The quadratic formula is a powerful tool for solving quadratic equations. It is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a$, $b$, and $c$ are the coefficients of the quadratic equation.

Q2: How do I apply the quadratic formula?

A2: To apply the quadratic formula, you need to substitute the values of $a$, $b$, and $c$ into the formula. Then, simplify the expression to get the final answer.

Q3: What is the difference between the quadratic formula and factoring?

A3: The quadratic formula and factoring are two different methods for solving quadratic equations. The quadratic formula is a general method that can be used to solve any quadratic equation, while factoring is a specific method that can be used to solve quadratic equations that can be factored.

Q4: Can I use the quadratic formula to solve quadratic equations with complex solutions?

A4: Yes, you can use the quadratic formula to solve quadratic equations with complex solutions. The quadratic formula will give you the complex solutions, and you can simplify them to get the final answer.

Q5: How do I determine the number of solutions to a quadratic equation?

A5: To determine the number of solutions to a quadratic equation, you need to look at the discriminant, which is the expression under the square root in the quadratic formula. If the discriminant is positive, the equation has two real solutions. If the discriminant is zero, the equation has one real solution. If the discriminant is negative, the equation has two complex solutions.

Q6: Can I use the quadratic formula to solve quadratic equations with rational coefficients?

A6: Yes, you can use the quadratic formula to solve quadratic equations with rational coefficients. The quadratic formula will give you the rational solutions, and you can simplify them to get the final answer.

Q7: How do I simplify the solutions to a quadratic equation?

A7: To simplify the solutions to a quadratic equation, you need to look at the numerator and the denominator of the expression. You can simplify the numerator and the denominator separately, and then simplify the expression as a whole.

Q8: Can I use the quadratic formula to solve quadratic equations with negative coefficients?

A8: Yes, you can use the quadratic formula to solve quadratic equations with negative coefficients. The quadratic formula will give you the solutions, and you can simplify them to get the final answer.

Q9: How do I determine the nature of the solutions to a quadratic equation?

A9: To determine the nature of the solutions to a quadratic equation, you need to look at the discriminant, which is the expression under the square root in the quadratic formula. 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.

Q10: Can I use the quadratic formula to solve quadratic equations with fractional coefficients?

A10: Yes, you can use the quadratic formula to solve quadratic equations with fractional coefficients. The quadratic formula will give you the solutions, and you can simplify them to get the final answer.

Conclusion

In this article, we have answered some of the frequently asked questions related to solving quadratic equations using the quadratic formula. We have covered topics such as the quadratic formula, applying the quadratic formula, and simplifying the solutions. We hope that this article has been helpful in clarifying any doubts you may have had about solving quadratic equations.

References

• "Algebra" by Michael Artin
• "Calculus" by Michael Spivak
• "Linear Algebra and Its Applications" by Gilbert Strang

Further Reading

• "Quadratic Equations" by Math Open Reference
• "Quadratic Formula" by Wolfram MathWorld
• "Solving Quadratic Equations" by Khan Academy