If $a$ And $b$ Are The Roots Of The Equation $3x^2 + 6x + 2 = 0$, Then Find:a) $a + B$b) $ab$c) $\frac{1}{a} + \frac{1}{b}$d) $\frac{3}{a+2} + \frac{3}{b+2}$e) $a^2 +

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Introduction

Quadratic equations are a fundamental concept in mathematics, and they have numerous applications in various fields, including physics, engineering, and economics. In this article, we will explore the roots of a quadratic equation and how to find various expressions involving these roots. We will use the quadratic equation 3x2+6x+2=03x^2 + 6x + 2 = 0 as a case study to demonstrate the concepts.

The Quadratic Formula

The quadratic formula is a powerful tool for finding the roots of a quadratic equation. It states that for an equation of the form ax2+bx+c=0ax^2 + bx + c = 0, the roots are given by:

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

In our case, the equation is 3x2+6x+2=03x^2 + 6x + 2 = 0, so we have a=3a = 3, b=6b = 6, and c=2c = 2. Plugging these values into the quadratic formula, we get:

x=−6±62−4(3)(2)2(3)x = \frac{-6 \pm \sqrt{6^2 - 4(3)(2)}}{2(3)}

Simplifying the expression under the square root, we get:

x=−6±36−246x = \frac{-6 \pm \sqrt{36 - 24}}{6}

x=−6±126x = \frac{-6 \pm \sqrt{12}}{6}

x=−6±236x = \frac{-6 \pm 2\sqrt{3}}{6}

x=−3±33x = \frac{-3 \pm \sqrt{3}}{3}

Therefore, the roots of the equation are x=−3+33x = \frac{-3 + \sqrt{3}}{3} and x=−3−33x = \frac{-3 - \sqrt{3}}{3}.

Finding the Sum and Product of the Roots

Now that we have found the roots of the equation, we can use them to find various expressions involving the roots. Let's start with the sum and product of the roots.

The sum of the roots is given by:

a+b=−3+33+−3−33a + b = \frac{-3 + \sqrt{3}}{3} + \frac{-3 - \sqrt{3}}{3}

Simplifying the expression, we get:

a+b=−63a + b = \frac{-6}{3}

a+b=−2a + b = -2

Therefore, the sum of the roots is −2-2.

The product of the roots is given by:

ab=−3+33×−3−33ab = \frac{-3 + \sqrt{3}}{3} \times \frac{-3 - \sqrt{3}}{3}

Simplifying the expression, we get:

ab=9−39ab = \frac{9 - 3}{9}

ab=69ab = \frac{6}{9}

ab=23ab = \frac{2}{3}

Therefore, the product of the roots is 23\frac{2}{3}.

Finding the Reciprocal of the Sum and Product of the Roots

Now, let's find the reciprocal of the sum and product of the roots.

The reciprocal of the sum of the roots is given by:

1a+b=1−2\frac{1}{a + b} = \frac{1}{-2}

Simplifying the expression, we get:

1a+b=−12\frac{1}{a + b} = -\frac{1}{2}

Therefore, the reciprocal of the sum of the roots is −12-\frac{1}{2}.

The reciprocal of the product of the roots is given by:

1ab=123\frac{1}{ab} = \frac{1}{\frac{2}{3}}

Simplifying the expression, we get:

1ab=32\frac{1}{ab} = \frac{3}{2}

Therefore, the reciprocal of the product of the roots is 32\frac{3}{2}.

Finding the Sum of the Reciprocals of the Roots

Now, let's find the sum of the reciprocals of the roots.

The sum of the reciprocals of the roots is given by:

1a+1b=1−3+33+1−3−33\frac{1}{a} + \frac{1}{b} = \frac{1}{\frac{-3 + \sqrt{3}}{3}} + \frac{1}{\frac{-3 - \sqrt{3}}{3}}

Simplifying the expression, we get:

1a+1b=3−3+3+3−3−3\frac{1}{a} + \frac{1}{b} = \frac{3}{-3 + \sqrt{3}} + \frac{3}{-3 - \sqrt{3}}

Using the conjugate to rationalize the denominators, we get:

1a+1b=3(−3−3)(−3+3)(−3−3)+3(−3+3)(−3−3)(−3+3)\frac{1}{a} + \frac{1}{b} = \frac{3(-3 - \sqrt{3})}{(-3 + \sqrt{3})(-3 - \sqrt{3})} + \frac{3(-3 + \sqrt{3})}{(-3 - \sqrt{3})(-3 + \sqrt{3})}

Simplifying the expression, we get:

1a+1b=3(−3−3)9−3+3(−3+3)9−3\frac{1}{a} + \frac{1}{b} = \frac{3(-3 - \sqrt{3})}{9 - 3} + \frac{3(-3 + \sqrt{3})}{9 - 3}

1a+1b=3(−3−3)+3(−3+3)6\frac{1}{a} + \frac{1}{b} = \frac{3(-3 - \sqrt{3}) + 3(-3 + \sqrt{3})}{6}

1a+1b=−186\frac{1}{a} + \frac{1}{b} = \frac{-18}{6}

1a+1b=−3\frac{1}{a} + \frac{1}{b} = -3

Therefore, the sum of the reciprocals of the roots is −3-3.

Finding the Sum of the Reciprocals of the Roots with a Constant Added

Now, let's find the sum of the reciprocals of the roots with a constant added.

The sum of the reciprocals of the roots with a constant added is given by:

3a+2+3b+2=3−3+33+2+3−3−33+2\frac{3}{a + 2} + \frac{3}{b + 2} = \frac{3}{\frac{-3 + \sqrt{3}}{3} + 2} + \frac{3}{\frac{-3 - \sqrt{3}}{3} + 2}

Simplifying the expression, we get:

3a+2+3b+2=3−3+3+63+3−3−3+63\frac{3}{a + 2} + \frac{3}{b + 2} = \frac{3}{\frac{-3 + \sqrt{3} + 6}{3}} + \frac{3}{\frac{-3 - \sqrt{3} + 6}{3}}

3a+2+3b+2=33+33+33−33\frac{3}{a + 2} + \frac{3}{b + 2} = \frac{3}{\frac{3 + \sqrt{3}}{3}} + \frac{3}{\frac{3 - \sqrt{3}}{3}}

Using the conjugate to rationalize the denominators, we get:

3a+2+3b+2=3(3−3)(3+3)(3−3)+3(3+3)(3−3)(3+3)\frac{3}{a + 2} + \frac{3}{b + 2} = \frac{3(3 - \sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})} + \frac{3(3 + \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})}

Simplifying the expression, we get:

3a+2+3b+2=3(3−3)9−3+3(3+3)9−3\frac{3}{a + 2} + \frac{3}{b + 2} = \frac{3(3 - \sqrt{3})}{9 - 3} + \frac{3(3 + \sqrt{3})}{9 - 3}

3a+2+3b+2=3(3−3)+3(3+3)6\frac{3}{a + 2} + \frac{3}{b + 2} = \frac{3(3 - \sqrt{3}) + 3(3 + \sqrt{3})}{6}

3a+2+3b+2=186\frac{3}{a + 2} + \frac{3}{b + 2} = \frac{18}{6}

3a+2+3b+2=3\frac{3}{a + 2} + \frac{3}{b + 2} = 3

Therefore, the sum of the reciprocals of the roots with a constant added is 33.

Conclusion

Introduction

Quadratic equations are a fundamental concept in mathematics, and they have numerous applications in various fields, including physics, engineering, and economics. In our previous article, we explored the roots of a quadratic equation and how to find various expressions involving these roots. In this article, we will answer some frequently asked questions about quadratic equations.

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. It is typically written in the form ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are constants, and xx is the variable.

Q: How do I solve a quadratic equation?

A: There are several methods to solve a quadratic equation, including factoring, using the quadratic formula, and completing the square. The quadratic formula is a powerful tool for finding the roots of a quadratic equation and is given by:

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

Q: What is the difference between the sum and product of the roots?

A: The sum of the roots is given by a+ba + b, and the product of the roots is given by abab. These two expressions are related to the coefficients of the quadratic equation and can be used to find the roots of the equation.

Q: How do I find the reciprocal of the sum and product of the roots?

A: The reciprocal of the sum of the roots is given by 1a+b\frac{1}{a + b}, and the reciprocal of the product of the roots is given by 1ab\frac{1}{ab}. These expressions can be used to find the roots of the equation.

Q: What is the sum of the reciprocals of the roots?

A: The sum of the reciprocals of the roots is given by 1a+1b\frac{1}{a} + \frac{1}{b}. This expression can be used to find the roots of the equation.

Q: How do I find the sum of the reciprocals of the roots with a constant added?

A: The sum of the reciprocals of the roots with a constant added is given by 3a+2+3b+2\frac{3}{a + 2} + \frac{3}{b + 2}. This expression can be used to find the roots of the equation.

Q: What are the applications of quadratic equations?

A: Quadratic equations have numerous applications in various fields, including physics, engineering, and economics. They are used to model real-world problems, such as the motion of objects, the growth of populations, and the behavior of electrical circuits.

Q: How do I use quadratic equations in real-world problems?

A: Quadratic equations can be used to model real-world problems by using the roots of the equation to find the solutions to the problem. For example, if you are designing a bridge, you can use a quadratic equation to find the height of the bridge that will allow it to span a certain distance.

Conclusion

In this article, we have answered some frequently asked questions about quadratic equations. We have covered topics such as solving quadratic equations, finding the sum and product of the roots, and using quadratic equations in real-world problems. We hope that this article has been helpful in understanding the concepts of quadratic equations.

Additional Resources

For more information on quadratic equations, we recommend the following resources:

  • Khan Academy: Quadratic Equations
  • Mathway: Quadratic Equations
  • Wolfram Alpha: Quadratic Equations

Final Thoughts

Quadratic equations are a fundamental concept in mathematics, and they have numerous applications in various fields. By understanding the concepts of quadratic equations, you can use them to model real-world problems and find solutions to complex problems. We hope that this article has been helpful in understanding the concepts of quadratic equations.