What Are The Solutions To The Quadratic Equation $3(x-4)^2=75$?A. $x=-9$ And $ X = 1 X=1 X = 1 [/tex] B. $x=-5$ And $x=5$ C. $ X = − 4 X=-4 X = − 4 [/tex] And $x=4$ D. $x=-1$ And

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Introduction to Quadratic Equations

Quadratic equations are a fundamental concept in mathematics, and they play a crucial role in various fields such as physics, engineering, and economics. 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 $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants, and $x$ is the variable.

Understanding the Given Quadratic Equation

The given quadratic equation is $3(x-4)^2=75$. To solve this equation, we need to isolate the variable $x$. The first step is to expand the squared term using the formula $(a-b)^2 = a^2 - 2ab + b^2$. Applying this formula to the given equation, we get:

3(x2−8x+16)=753(x^2 - 8x + 16) = 75

Expanding and Simplifying the Equation

Now, let's expand and simplify the equation:

3x2−24x+48=753x^2 - 24x + 48 = 75

Subtracting 75 from both sides of the equation, we get:

3x2−24x−27=03x^2 - 24x - 27 = 0

Solving the Quadratic Equation

To solve the quadratic equation, we can use the quadratic formula, which is:

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

In this case, $a = 3$, $b = -24$, and $c = -27$. Plugging these values into the quadratic formula, we get:

x=−(−24)±(−24)2−4(3)(−27)2(3)x = \frac{-(-24) \pm \sqrt{(-24)^2 - 4(3)(-27)}}{2(3)}

Simplifying the Quadratic Formula

Now, let's simplify the quadratic formula:

x=24±576+3246x = \frac{24 \pm \sqrt{576 + 324}}{6}

x=24±9006x = \frac{24 \pm \sqrt{900}}{6}

x=24±306x = \frac{24 \pm 30}{6}

Finding the Solutions

Now, let's find the solutions to the quadratic equation:

x=24+306x = \frac{24 + 30}{6}

x=546x = \frac{54}{6}

x=9x = 9

x=24−306x = \frac{24 - 30}{6}

x=−66x = \frac{-6}{6}

x=−1x = -1

Conclusion

In conclusion, the solutions to the quadratic equation $3(x-4)^2=75$ are $x = 9$ and $x = -1$. These solutions can be verified by plugging them back into the original equation.

Final Answer

The final answer is: 9\boxed{9} and −1\boxed{-1}

Introduction

Quadratic equations are a fundamental concept in mathematics, and they play a crucial role in various fields such as physics, engineering, and economics. In our previous article, we discussed the solutions to the quadratic equation $3(x-4)^2=75$. In this article, we will provide a comprehensive Q&A guide to help you understand quadratic equations and their solutions.

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 $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants, and $x$ is the variable.

Q: How do I solve a quadratic equation?

A: There are several methods to solve a quadratic equation, including:

  • Factoring: If the quadratic expression can be factored into the product of two binomials, you can solve the equation by setting each factor equal to zero.
  • Quadratic formula: The quadratic formula is a general method for solving quadratic equations. It is given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
  • Graphing: You can also solve a quadratic equation by graphing the related function and finding the x-intercepts.

Q: What is the quadratic formula?

A: The quadratic formula is a general method for solving quadratic equations. It is given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Q: How do I use the quadratic formula?

A: To use the quadratic formula, you need to identify the values of $a$, $b$, and $c$ in the quadratic equation. Then, you can plug these values into the quadratic formula and simplify to find the solutions.

Q: What are the solutions to the quadratic equation $x^2 + 4x + 4 = 0$?

A: To solve the quadratic equation $x^2 + 4x + 4 = 0$, we can use the quadratic formula:

x=−4±42−4(1)(4)2(1)x = \frac{-4 \pm \sqrt{4^2 - 4(1)(4)}}{2(1)}

x=−4±16−162x = \frac{-4 \pm \sqrt{16 - 16}}{2}

x=−4±02x = \frac{-4 \pm \sqrt{0}}{2}

x=−42x = \frac{-4}{2}

x=−2x = -2

Q: What are the solutions to the quadratic equation $x^2 - 6x + 8 = 0$?

A: To solve the quadratic equation $x^2 - 6x + 8 = 0$, we can use the quadratic formula:

x=−(−6)±(−6)2−4(1)(8)2(1)x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(8)}}{2(1)}

x=6±36−322x = \frac{6 \pm \sqrt{36 - 32}}{2}

x=6±42x = \frac{6 \pm \sqrt{4}}{2}

x=6±22x = \frac{6 \pm 2}{2}

x=6+22x = \frac{6 + 2}{2}

x=82x = \frac{8}{2}

x=4x = 4

x=6−22x = \frac{6 - 2}{2}

x=42x = \frac{4}{2}

x=2x = 2

Q: What are the solutions to the quadratic equation $3(x-4)^2=75$?

A: In our previous article, we discussed the solutions to the quadratic equation $3(x-4)^2=75$. The solutions are $x = 9$ and $x = -1$.

Conclusion

In conclusion, quadratic equations are a fundamental concept in mathematics, and they play a crucial role in various fields such as physics, engineering, and economics. We hope that this Q&A guide has helped you understand quadratic equations and their solutions.

Final Answer

The final answer is: 9\boxed{9} and −1\boxed{-1}