What Are The Roots Of The Equation?$9x - 2 = -5x^2$A. $\frac{1}{5}$ And -2 B. $\frac{-9 \pm \sqrt{41}}{10}$C. 2 And $-\frac{1}{5}$D. $\frac{9 \pm \sqrt{41}}{10}$

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Introduction

In mathematics, solving equations is a fundamental concept that helps us find the value of unknown variables. When it comes to quadratic equations, we often encounter expressions in the form of ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are constants. However, in this article, we will delve into a slightly different scenario, where the equation is given as 9xβˆ’2=βˆ’5x29x - 2 = -5x^2. Our primary objective is to find the roots of this equation, which will lead us to the values of xx that satisfy the given equation.

Understanding the Equation

Before we proceed with solving the equation, let's first understand its structure. The given equation is 9xβˆ’2=βˆ’5x29x - 2 = -5x^2. To make it more manageable, we can rewrite it in the standard form of a quadratic equation, which is ax2+bx+c=0ax^2 + bx + c = 0. By rearranging the terms, we get 5x2+9xβˆ’2=05x^2 + 9x - 2 = 0. Now, we have a quadratic equation in the form of ax2+bx+c=0ax^2 + bx + c = 0, where a=5a = 5, b=9b = 9, and c=βˆ’2c = -2.

Solving the Quadratic Equation

To find the roots of the quadratic equation 5x2+9xβˆ’2=05x^2 + 9x - 2 = 0, we can use the quadratic formula, which is given by:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In this case, a=5a = 5, b=9b = 9, and c=βˆ’2c = -2. Plugging these values into the quadratic formula, we get:

x=βˆ’9Β±92βˆ’4(5)(βˆ’2)2(5)x = \frac{-9 \pm \sqrt{9^2 - 4(5)(-2)}}{2(5)}

Simplifying the expression under the square root, we get:

x=βˆ’9Β±81+4010x = \frac{-9 \pm \sqrt{81 + 40}}{10}

x=βˆ’9Β±12110x = \frac{-9 \pm \sqrt{121}}{10}

x=βˆ’9Β±1110x = \frac{-9 \pm 11}{10}

Now, we have two possible values for xx, which are:

x=βˆ’9+1110x = \frac{-9 + 11}{10} and x=βˆ’9βˆ’1110x = \frac{-9 - 11}{10}

Simplifying these expressions, we get:

x=210x = \frac{2}{10} and x=βˆ’2010x = \frac{-20}{10}

x=15x = \frac{1}{5} and x=βˆ’2x = -2

Conclusion

In conclusion, the roots of the equation 9xβˆ’2=βˆ’5x29x - 2 = -5x^2 are 15\frac{1}{5} and βˆ’2-2. These values satisfy the given equation, and they can be obtained by using the quadratic formula. The quadratic formula is a powerful tool for solving quadratic equations, and it helps us find the roots of the equation in a straightforward manner.

Discussion

The quadratic formula is a fundamental concept in mathematics, and it has numerous applications in various fields, such as physics, engineering, and economics. In this article, we have used the quadratic formula to find the roots of the equation 9xβˆ’2=βˆ’5x29x - 2 = -5x^2. The quadratic formula is a powerful tool for solving quadratic equations, and it helps us find the roots of the equation in a straightforward manner.

Final Answer

The final answer is 15Β andΒ βˆ’2\boxed{\frac{1}{5} \text{ and } -2}.

Comparison of Options

Let's compare the options given in the problem with the actual roots of the equation.

Option A: 15\frac{1}{5} and βˆ’2-2 Option B: βˆ’9Β±4110\frac{-9 \pm \sqrt{41}}{10} Option C: 22 and βˆ’15-\frac{1}{5} Option D: 9Β±4110\frac{9 \pm \sqrt{41}}{10}

The correct answer is Option A: 15\frac{1}{5} and βˆ’2-2.

Comparison with Other Methods

We can also solve the quadratic equation using other methods, such as factoring or completing the square. However, the quadratic formula is a more general method that can be applied to any quadratic equation.

Conclusion

In conclusion, the roots of the equation 9xβˆ’2=βˆ’5x29x - 2 = -5x^2 are 15\frac{1}{5} and βˆ’2-2. These values satisfy the given equation, and they can be obtained by using the quadratic formula. The quadratic formula is a powerful tool for solving quadratic equations, and it helps us find the roots of the equation in a straightforward manner.

Introduction

In our previous article, we explored the roots of the equation 9xβˆ’2=βˆ’5x29x - 2 = -5x^2. We used the quadratic formula to find the values of xx that satisfy the given equation. In this article, we will provide a Q&A section to help clarify any doubts or questions you may have about the roots of the equation.

Q&A Section

Q: What is the quadratic formula?

A: The quadratic formula is a mathematical formula that helps us find the roots of a quadratic equation. It is given by:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Q: How do I use the quadratic formula to find the roots of the equation?

A: To use the quadratic formula, you need to identify the values of aa, bb, and cc in the quadratic equation. Then, you can plug these values into the quadratic formula to find the roots of the equation.

Q: What are the roots of the equation 9xβˆ’2=βˆ’5x29x - 2 = -5x^2?

A: The roots of the equation 9xβˆ’2=βˆ’5x29x - 2 = -5x^2 are 15\frac{1}{5} and βˆ’2-2.

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 calculate the value of b2βˆ’4acb^2 - 4ac. Then, you can take the square root of this value to find the roots of the equation.

Q: Can I use other methods to solve the quadratic equation?

A: Yes, you can use other methods such as factoring or completing the square to solve the quadratic equation. However, the quadratic formula is a more general method that can be applied to any quadratic equation.

Q: What are the applications of the quadratic formula?

A: The quadratic formula has numerous applications in various fields such as physics, engineering, and economics. It helps us find the roots of quadratic equations, which is essential in solving problems in these fields.

Q: Can I use the quadratic formula to solve quadratic equations with complex roots?

A: Yes, you can use the quadratic formula to solve quadratic equations with complex roots. However, you need to be careful when simplifying the expression under the square root, as it may involve complex numbers.

Q: How do I determine the nature of the roots of a quadratic equation?

A: To determine the nature of the roots of a quadratic equation, you need to calculate the discriminant, which is given by b2βˆ’4acb^2 - 4ac. If the discriminant is positive, the roots are real and distinct. If the discriminant is zero, the roots are real and equal. If the discriminant is negative, the roots are complex.

Conclusion

In conclusion, the quadratic formula is a powerful tool for solving quadratic equations. It helps us find the roots of the equation in a straightforward manner. We hope that this Q&A article has helped clarify any doubts or questions you may have about the roots of the equation 9xβˆ’2=βˆ’5x29x - 2 = -5x^2. If you have any further questions, please feel free to ask.

Final Answer

The final answer is 15Β andΒ βˆ’2\boxed{\frac{1}{5} \text{ and } -2}.

Comparison of Options

Let's compare the options given in the problem with the actual roots of the equation.

Option A: 15\frac{1}{5} and βˆ’2-2 Option B: βˆ’9Β±4110\frac{-9 \pm \sqrt{41}}{10} Option C: 22 and βˆ’15-\frac{1}{5} Option D: 9Β±4110\frac{9 \pm \sqrt{41}}{10}

The correct answer is Option A: 15\frac{1}{5} and βˆ’2-2.

Comparison with Other Methods

We can also solve the quadratic equation using other methods, such as factoring or completing the square. However, the quadratic formula is a more general method that can be applied to any quadratic equation.

Conclusion

In conclusion, the roots of the equation 9xβˆ’2=βˆ’5x29x - 2 = -5x^2 are 15\frac{1}{5} and βˆ’2-2. These values satisfy the given equation, and they can be obtained by using the quadratic formula. The quadratic formula is a powerful tool for solving quadratic equations, and it helps us find the roots of the equation in a straightforward manner.