Verify That $k = \frac{-3 + 3\sqrt{3}}{2}$.$\[ \begin{array}{rl} (2k + 3)^2 - 24 & = 3 \\ {[2(k) + 3]^2 - 24} & \stackrel{?}{=} 3 \\ {[2 \left(\frac{-3 + 3\sqrt{3}}{2}\right) + 3]^2 - 24} & \stackrel{?}{=} 3 \\ {[-3 + 3\sqrt{3} + 3]^2 -

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Introduction

In mathematics, verifying the value of a variable in a given expression is an essential skill that helps in solving problems and understanding mathematical concepts. In this article, we will focus on verifying the value of $k$ in the expression $(2k + 3)^2 - 24 = 3$. We will use algebraic manipulations to simplify the expression and check if the value of $k$ satisfies the given equation.

The Given Equation

The given equation is $(2k + 3)^2 - 24 = 3$. Our goal is to verify the value of $k$ in this equation. To do this, we will start by simplifying the left-hand side of the equation.

Simplifying the Left-Hand Side

We can simplify the left-hand side of the equation by expanding the square and combining like terms.

(2k + 3)^2 - 24 = 3
(2k + 3)(2k + 3) - 24 = 3
4k^2 + 12k + 9 - 24 = 3
4k^2 + 12k - 15 = 3

Isolating the Variable

Now, we can isolate the variable $k$ by moving all the terms to one side of the equation.

4k^2 + 12k - 15 = 3
4k^2 + 12k - 18 = 0

Factoring the Quadratic Expression

We can factor the quadratic expression on the left-hand side of the equation.

4k^2 + 12k - 18 = 0
2(2k^2 + 6k - 9) = 0
2(2k - 3)(k + 3) = 0

Finding the Values of $k$

Now, we can find the values of $k$ by setting each factor equal to zero.

2(2k - 3)(k + 3) = 0
2k - 3 = 0 \quad \text{or} \quad k + 3 = 0
2k = 3 \quad \text{or} \quad k = -3
k = \frac{3}{2} \quad \text{or} \quad k = -3

Verifying the Value of $k$

We are given that $k = \frac{-3 + 3\sqrt{3}}{2}$. We need to verify if this value satisfies the given equation.

(2k + 3)^2 - 24 = 3
(2 \left(\frac{-3 + 3\sqrt{3}}{2}\right) + 3)^2 - 24 = 3
(-3 + 3\sqrt{3} + 3)^2 - 24 = 3
(3\sqrt{3})^2 - 24 = 3
27 - 24 = 3
3 = 3

As we can see, the value of $k = \frac{-3 + 3\sqrt{3}}{2}$ satisfies the given equation.

Conclusion

In this article, we verified the value of $k$ in the expression $(2k + 3)^2 - 24 = 3$. We used algebraic manipulations to simplify the expression and check if the value of $k$ satisfies the given equation. We found that the value of $k = \frac{-3 + 3\sqrt{3}}{2}$ satisfies the given equation.

References

• [1] "Algebra" by Michael Artin
• [2] "Calculus" by Michael Spivak

Discussion

The value of $k = \frac{-3 + 3\sqrt{3}}{2}$ is a complex number. It can be written in the form $a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit.

k = \frac{-3 + 3\sqrt{3}}{2}
k = \frac{-3 + 3\sqrt{3}}{2} + 0i
k = -\frac{3}{2} + \frac{3\sqrt{3}}{2}i

The value of $k$ can be visualized as a point on the complex plane. The real part of $k$ is $-\frac{3}{2}$ and the imaginary part of $k$ is $\frac{3\sqrt{3}}{2}$.

\text{Real part of } k = -\frac{3}{2}
\text{Imaginary part of } k = \frac{3\sqrt{3}}{2}

The value of $k$ can be used to solve problems in mathematics and science. For example, it can be used to find the roots of a quadratic equation.

ax^2 + bx + c = 0
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The value of $k$ can also be used to find the solutions to a system of linear equations.

\begin{align*}
ax + by &= c \\
dx + ey &= f
\end{align*}
\begin{align*}
x &= \frac{cf - bd}{ad - be} \\
y &= \frac{af - cd}{ad - be}
\end{align*}

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Introduction

In our previous article, we verified the value of $k$ in the expression $(2k + 3)^2 - 24 = 3$. We used algebraic manipulations to simplify the expression and check if the value of $k$ satisfies the given equation. In this article, we will answer some frequently asked questions about the value of $k$ and its applications.

Q: What is the value of $k$?

A: The value of $k$ is $\frac{-3 + 3\sqrt{3}}{2}$.

Q: How did you verify the value of $k$?

A: We used algebraic manipulations to simplify the expression $(2k + 3)^2 - 24 = 3$ and check if the value of $k$ satisfies the given equation.

Q: What is the significance of the value of $k$?

A: The value of $k$ is an important concept in mathematics and science. It can be used to solve problems in many areas, including algebra, geometry, and calculus.

Q: Can you give an example of how to use the value of $k$ to solve a problem?

A: Yes, here is an example:

Suppose we want to find the roots of the quadratic equation $x^2 + 4x + 4 = 0$. We can use the value of $k$ to find the roots.

x^2 + 4x + 4 = 0
x^2 + 2(2)x + 2^2 = 0
(x + 2)^2 = 0
x + 2 = 0
x = -2

Q: Can you explain the concept of complex numbers?

A: Yes, complex numbers are numbers that have both real and imaginary parts. They can be written in the form $a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit.

k = \frac{-3 + 3\sqrt{3}}{2}
k = \frac{-3 + 3\sqrt{3}}{2} + 0i
k = -\frac{3}{2} + \frac{3\sqrt{3}}{2}i

Q: How do you add and subtract complex numbers?

A: To add and subtract complex numbers, we add and subtract the real parts and the imaginary parts separately.

(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi) - (c + di) = (a - c) + (b - d)i

Q: Can you explain the concept of the complex plane?

A: Yes, the complex plane is a two-dimensional plane that represents complex numbers. The real part of a complex number is represented on the x-axis, and the imaginary part is represented on the y-axis.

\text{Real part of } k = -\frac{3}{2}
\text{Imaginary part of } k = \frac{3\sqrt{3}}{2}

Q: How do you multiply complex numbers?

A: To multiply complex numbers, we use the distributive property and the fact that $i^2 = -1$.

(a + bi)(c + di) = (ac - bd) + (ad + bc)i

Conclusion

In this article, we answered some frequently asked questions about the value of $k$ and its applications. We explained the concept of complex numbers, the complex plane, and how to add, subtract, and multiply complex numbers. We also gave an example of how to use the value of $k$ to solve a problem.

References

• [1] "Algebra" by Michael Artin
• [2] "Calculus" by Michael Spivak
• [3] "Complex Analysis" by Serge Lang

Discussion

The value of $k$ is an important concept in mathematics and science. It can be used to solve problems in many areas, including algebra, geometry, and calculus. The concept of complex numbers and the complex plane are also important in many areas of mathematics and science.

\text{Complex numbers are numbers that have both real and imaginary parts.}
\text{The complex plane is a two-dimensional plane that represents complex numbers.}

The value of $k$ can be used to solve problems in many areas of mathematics and science. It is an important concept that can be used to solve a wide range of problems.

\text{The value of } k \text{ is an important concept in mathematics and science.}
\text{It can be used to solve problems in many areas, including algebra, geometry, and calculus.}