Which Shows The Equation Below Written In The Form $a X^2 + B X + C = 0$?$x + 9 = 4(x - 1)^2$A. $4 X^2 - 7 X + 13 = 0$B. $4 X^2 - 9 X + 13 = 0$C. $4 X^2 - 7 X - 5 = 0$D. $4 X^2 - 9 X - 5 = 0$

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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 this article, we will explore how to solve quadratic equations in the form ax2+bx+c=0ax^2 + bx + c = 0. We will also provide a step-by-step guide on how to rewrite a given equation in this form.

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, and xx is the variable.

Rewriting the Equation

The given equation is x+9=4(x−1)2x + 9 = 4(x - 1)^2. To rewrite this equation in the form ax2+bx+c=0ax^2 + bx + c = 0, we need to expand the right-hand side of the equation.

Expanding the Right-Hand Side

To expand the right-hand side of the equation, we need to use the binomial expansion formula. The binomial expansion formula is given by:

(a+b)n=(n0)anb0+(n1)an−1b1+(n2)an−2b2+...+(nn−1)a1bn−1+(nn)a0bn(a + b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + ... + \binom{n}{n-1} a^1 b^{n-1} + \binom{n}{n} a^0 b^n

In this case, we have (x−1)2(x - 1)^2, which can be expanded as:

(x−1)2=x2−2x+1(x - 1)^2 = x^2 - 2x + 1

Now, we can substitute this expression back into the original equation:

x+9=4(x2−2x+1)x + 9 = 4(x^2 - 2x + 1)

Distributing the 4

To distribute the 4, we need to multiply each term inside the parentheses by 4:

x+9=4x2−8x+4x + 9 = 4x^2 - 8x + 4

Rearranging the Terms

To rewrite the equation in the form ax2+bx+c=0ax^2 + bx + c = 0, we need to rearrange the terms. We can do this by subtracting xx from both sides of the equation and subtracting 9 from both sides of the equation:

0=4x2−9x−50 = 4x^2 - 9x - 5

Conclusion

In conclusion, the given equation x+9=4(x−1)2x + 9 = 4(x - 1)^2 can be rewritten in the form ax2+bx+c=0ax^2 + bx + c = 0 as:

0=4x2−9x−50 = 4x^2 - 9x - 5

Therefore, the correct answer is:

D. 4x2−9x−5=04 x^2 - 9 x - 5 = 0

Example Problems

Problem 1

Solve the equation x+2=3(x−1)2x + 2 = 3(x - 1)^2 in the form ax2+bx+c=0ax^2 + bx + c = 0.

Solution

To solve this problem, we need to follow the same steps as before. We can start by expanding the right-hand side of the equation:

(x−1)2=x2−2x+1(x - 1)^2 = x^2 - 2x + 1

Now, we can substitute this expression back into the original equation:

x+2=3(x2−2x+1)x + 2 = 3(x^2 - 2x + 1)

Distributing the 3, we get:

x+2=3x2−6x+3x + 2 = 3x^2 - 6x + 3

Rearranging the terms, we get:

0=3x2−7x+10 = 3x^2 - 7x + 1

Therefore, the correct answer is:

A. 3x2−7x+1=03 x^2 - 7 x + 1 = 0

Problem 2

Solve the equation x−1=2(x+2)2x - 1 = 2(x + 2)^2 in the form ax2+bx+c=0ax^2 + bx + c = 0.

Solution

To solve this problem, we need to follow the same steps as before. We can start by expanding the right-hand side of the equation:

(x+2)2=x2+4x+4(x + 2)^2 = x^2 + 4x + 4

Now, we can substitute this expression back into the original equation:

x−1=2(x2+4x+4)x - 1 = 2(x^2 + 4x + 4)

Distributing the 2, we get:

x−1=2x2+8x+8x - 1 = 2x^2 + 8x + 8

Rearranging the terms, we get:

0=2x2+7x+90 = 2x^2 + 7x + 9

Therefore, the correct answer is:

B. 2x2+7x+9=02 x^2 + 7 x + 9 = 0

Conclusion

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 this article, we will provide a Q&A guide on quadratic equations, covering topics such as solving quadratic equations, graphing quadratic functions, and more.

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, and xx is the variable.

Q: How do I solve a quadratic equation?

A: To solve a quadratic equation, you can use the following steps:

  1. Expand the right-hand side of the equation, if necessary.
  2. Distribute the coefficients, if necessary.
  3. Rearrange the terms to get the equation in the form ax2+bx+c=0ax^2 + bx + c = 0.
  4. Use the quadratic formula to find the solutions: 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 used to find the solutions of a quadratic equation. It is given by:

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

Q: How do I graph a quadratic function?

A: To graph a quadratic function, you can use the following steps:

  1. Determine the vertex of the parabola, which is the point where the parabola changes direction.
  2. Determine the x-intercepts of the parabola, which are the points where the parabola crosses the x-axis.
  3. Plot the vertex and the x-intercepts on a coordinate plane.
  4. Draw a smooth curve through the points to create the graph of the quadratic function.

Q: What is the difference between a quadratic equation and a quadratic function?

A: A quadratic equation is an equation that can be written in the form ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are constants, and xx is the variable. A quadratic function, on the other hand, is a function that can be written in the form f(x)=ax2+bx+cf(x) = ax^2 + bx + c, where aa, bb, and cc are constants, and xx is the variable.

Q: Can I use a calculator to solve a quadratic equation?

A: Yes, you can use a calculator to solve a quadratic equation. Most calculators have a built-in quadratic formula function that you can use to find the solutions of a quadratic equation.

Q: What are some common mistakes to avoid when solving quadratic equations?

A: Some common mistakes to avoid when solving quadratic equations include:

  • Not expanding the right-hand side of the equation, if necessary.
  • Not distributing the coefficients, if necessary.
  • Not rearranging the terms to get the equation in the form ax2+bx+c=0ax^2 + bx + c = 0.
  • Not using the quadratic formula correctly.

Q: Can I use quadratic equations in real-world applications?

A: Yes, quadratic equations have many real-world applications, including:

  • Physics: Quadratic equations are used to model the motion of objects under the influence of gravity.
  • Engineering: Quadratic equations are used to design and optimize systems, such as bridges and buildings.
  • Economics: Quadratic equations are used to model the behavior of economic systems, such as supply and demand.

Conclusion

In conclusion, quadratic equations are a fundamental concept in mathematics, and they have many real-world applications. By understanding how to solve quadratic equations, graph quadratic functions, and use the quadratic formula, you can apply quadratic equations to a wide range of problems and situations.