Solve For X X X In The Equation X 2 − 12 X + 59 = 0 X^2 - 12x + 59 = 0 X 2 − 12 X + 59 = 0 .A. X = − 12 ± 85 X = -12 \pm \sqrt{85} X = − 12 ± 85 ​ B. X = − 6 ± 23 I X = -6 \pm \sqrt{23} I X = − 6 ± 23 ​ I C. X = 6 ± 23 I X = 6 \pm \sqrt{23} I X = 6 ± 23 ​ I D. X = 12 ± 85 X = 12 \pm \sqrt{85} X = 12 ± 85 ​

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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, x2−12x+59=0x^2 - 12x + 59 = 0, 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 equation, x2−12x+59=0x^2 - 12x + 59 = 0, the coefficients are a=1a = 1, b=−12b = -12, and c=59c = 59.

The Quadratic Formula

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

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

In our equation, x2−12x+59=0x^2 - 12x + 59 = 0, we can plug in the values of aa, bb, and cc into the quadratic formula:

x=−(−12)±(−12)2−4(1)(59)2(1)x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(59)}}{2(1)}

Simplifying the Quadratic Formula

Let's simplify the expression under the square root:

(−12)2−4(1)(59)=144−236=−92(-12)^2 - 4(1)(59) = 144 - 236 = -92

Since the expression under the square root is negative, we know that the solutions will be complex numbers.

Complex Solutions

The quadratic formula now becomes:

x=12±−922x = \frac{12 \pm \sqrt{-92}}{2}

We can simplify the expression under the square root by factoring out a negative sign:

−92=−1⋅92=i92\sqrt{-92} = \sqrt{-1} \cdot \sqrt{92} = i\sqrt{92}

where ii is the imaginary unit, which satisfies i2=−1i^2 = -1.

Simplifying the Complex Solutions

Now we can simplify the complex solutions:

x=12±i922=6±i922x = \frac{12 \pm i\sqrt{92}}{2} = 6 \pm \frac{i\sqrt{92}}{2}

Simplifying the Imaginary Part

We can simplify the imaginary part by factoring out a square root:

i922=i4⋅232=i4⋅232=2i232=i23\frac{i\sqrt{92}}{2} = \frac{i\sqrt{4 \cdot 23}}{2} = \frac{i\sqrt{4} \cdot \sqrt{23}}{2} = \frac{2i\sqrt{23}}{2} = i\sqrt{23}

Final Solutions

The final solutions are:

x=6±i23x = 6 \pm i\sqrt{23}

Conclusion

In this article, we solved the quadratic equation x2−12x+59=0x^2 - 12x + 59 = 0 using the quadratic formula. We found that the solutions are complex numbers, which can be written in the form x=6±i23x = 6 \pm i\sqrt{23}. This demonstrates the importance of the quadratic formula in solving quadratic equations and highlights the need to consider complex solutions when working with quadratic equations.

Answer

The correct answer is:

C. x=6±23ix = 6 \pm \sqrt{23} i

Note: The original equation x2−12x+59=0x^2 - 12x + 59 = 0 does not have real solutions, and the correct answer is a complex number.

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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 our previous article, we solved the quadratic equation x2−12x+59=0x^2 - 12x + 59 = 0 using the quadratic formula. In this article, we will answer some frequently asked questions about quadratic equations and provide additional insights and examples.

Q&A

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. The general form of a quadratic equation is ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are constants.

Q: How do I solve a quadratic equation?

A: There are several methods to solve a quadratic equation, including factoring, the quadratic formula, and graphing. The quadratic formula is a powerful tool for solving quadratic equations and is given by:

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

Q: What is the quadratic formula?

A: The quadratic formula is a formula for solving quadratic equations of the form ax2+bx+c=0ax^2 + bx + c = 0. It is given by:

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

Q: When do I use the quadratic formula?

A: You use the quadratic formula when the quadratic equation cannot be factored easily or when you need to find the solutions quickly.

Q: What is the difference between real and complex solutions?

A: Real solutions are solutions that are real numbers, while complex solutions are solutions that are complex numbers, which have both real and imaginary parts.

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

A: You can determine if a quadratic equation has real or complex solutions by looking 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.

Q: What is the discriminant?

A: The discriminant is the expression under the square root in the quadratic formula, which is given by b2−4acb^2 - 4ac.

Q: How do I simplify complex solutions?

A: To simplify complex solutions, you can use the following steps:

  1. Factor out a negative sign from the expression under the square root.
  2. Simplify the expression under the square root.
  3. Simplify the complex solution by combining the real and imaginary parts.

Examples

Example 1: Solving a Quadratic Equation with Real Solutions

Solve the quadratic equation x2+5x+6=0x^2 + 5x + 6 = 0.

A: Using the quadratic formula, we get:

x=−5±52−4(1)(6)2(1)x = \frac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)}

Simplifying the expression under the square root, we get:

x=−5±25−242=−5±12=−5±12x = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm \sqrt{1}}{2} = \frac{-5 \pm 1}{2}

Therefore, the solutions are:

x=−5+12=−2x = \frac{-5 + 1}{2} = -2

x=−5−12=−3x = \frac{-5 - 1}{2} = -3

Example 2: Solving a Quadratic Equation with Complex Solutions

Solve the quadratic equation x2−12x+59=0x^2 - 12x + 59 = 0.

A: Using the quadratic formula, we get:

x=−(−12)±(−12)2−4(1)(59)2(1)x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(59)}}{2(1)}

Simplifying the expression under the square root, we get:

x=12±−922=12±i922x = \frac{12 \pm \sqrt{-92}}{2} = \frac{12 \pm i\sqrt{92}}{2}

Simplifying the complex solution, we get:

x=6±i23x = 6 \pm i\sqrt{23}

Conclusion

In this article, we answered some frequently asked questions about quadratic equations and provided additional insights and examples. We also solved two quadratic equations, one with real solutions and one with complex solutions. We hope this article has been helpful in understanding quadratic equations and how to solve them.