Given That $x^2 - 6x + 1 = (x-a)^2 - B$ For All Values Of $x$,(i) Find The Value Of \$a$[/tex\] And The Value Of $b$.

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Solving the Quadratic Equation: Finding the Value of a and b

In this article, we will delve into the world of quadratic equations and explore the process of solving a given equation to find the values of its coefficients. The equation in question is $x^2 - 6x + 1 = (x-a)^2 - b$, and our goal is to determine the values of $a$ and $b$ for all values of $x$. We will use algebraic manipulation and comparison of coefficients to arrive at the solution.

Expanding the Right-Hand Side

To begin, we expand the right-hand side of the equation using the binomial expansion formula:

(x−a)2−b=x2−2ax+a2−b(x-a)^2 - b = x^2 - 2ax + a^2 - b

Now, we can rewrite the original equation as:

x2−6x+1=x2−2ax+a2−bx^2 - 6x + 1 = x^2 - 2ax + a^2 - b

Comparing Coefficients

We can now compare the coefficients of the corresponding terms on both sides of the equation. This will allow us to form a system of equations that we can solve to find the values of $a$ and $b$.

  • The coefficient of $x^2$ is the same on both sides, so we have no equation to form from this comparison.

  • The coefficient of $x$ on the left-hand side is $-6$, and on the right-hand side, it is $-2a$. Therefore, we have the equation:

    −6=−2a-6 = -2a

  • The constant term on the left-hand side is $1$, and on the right-hand side, it is $a^2 - b$. Therefore, we have the equation:

    1=a2−b1 = a^2 - b

Solving for a

We can now solve the equation $-6 = -2a$ for $a$ by dividing both sides by $-2$:

a=−6−2=3a = \frac{-6}{-2} = 3

Substituting a into the Second Equation

Now that we have found the value of $a$, we can substitute it into the second equation:

1=a2−b1 = a^2 - b

1=32−b1 = 3^2 - b

1=9−b1 = 9 - b

Solving for b

We can now solve the equation $1 = 9 - b$ for $b$ by subtracting $9$ from both sides and then multiplying both sides by $-1$:

−8=−b-8 = -b

b=8b = 8

In this article, we have solved the quadratic equation $x^2 - 6x + 1 = (x-a)^2 - b$ to find the values of $a$ and $b$. We used algebraic manipulation and comparison of coefficients to arrive at the solution. The value of $a$ is $3$, and the value of $b$ is $8$. These values are valid for all values of $x$.

The final answer is:

  • a=3a = 3

  • b = 8$<br/>

Quadratic Equation Q&A: Understanding the Solution

In our previous article, we solved the quadratic equation $x^2 - 6x + 1 = (x-a)^2 - b$ to find the values of $a$ and $b$. We used algebraic manipulation and comparison of coefficients to arrive at the solution. In this article, we will address some common questions and concerns related to the solution.

Q: What is the significance of the value of a?

A: The value of $a$ represents the horizontal shift of the parabola. In this case, the value of $a$ is $3$, which means that the parabola is shifted $3$ units to the right.

Q: How does the value of b affect the parabola?

A: The value of $b$ represents the vertical shift of the parabola. In this case, the value of $b$ is $8$, which means that the parabola is shifted $8$ units upwards.

Q: Can you explain the process of comparing coefficients?

A: Comparing coefficients involves matching the coefficients of the corresponding terms on both sides of the equation. In this case, we compared the coefficients of the $x$ terms and the constant terms to form a system of equations.

Q: What is the difference between the original equation and the expanded equation?

A: The original equation is $x^2 - 6x + 1 = (x-a)^2 - b$, while the expanded equation is $x^2 - 2ax + a^2 - b$. The expanded equation is obtained by expanding the right-hand side of the original equation using the binomial expansion formula.

Q: Can you provide an example of how to use the values of a and b in a real-world scenario?

A: Suppose we are designing a parabolic arch with a height of $8$ units and a base length of $6$ units. We can use the values of $a$ and $b$ to determine the equation of the parabola and calculate the coordinates of the vertex.

Q: What are some common applications of quadratic equations?

A: Quadratic equations have numerous applications in various fields, including physics, engineering, economics, and computer science. Some common applications include:

  • Modeling the trajectory of a projectile
  • Determining the maximum or minimum value of a function
  • Solving optimization problems
  • Analyzing the behavior of electrical circuits

In this article, we have addressed some common questions and concerns related to the solution of the quadratic equation $x^2 - 6x + 1 = (x-a)^2 - b$. We have explained the significance of the values of $a$ and $b$, the process of comparing coefficients, and some common applications of quadratic equations. We hope that this article has provided a deeper understanding of the solution and its implications.

The final answer is:

  • a=3a = 3

  • b=8b = 8