Find The Values Of $a$ And $b$ Such That ${x^2 + 2x - 7 = (x + A)^2 + B}$

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Introduction

In this article, we will delve into the world of quadratic equations and explore a method to find the values of aa and bb in the given equation. The equation in question is x2+2xβˆ’7=(x+a)2+bx^2 + 2x - 7 = (x + a)^2 + b. Our goal is to determine the values of aa and bb that satisfy this equation.

Understanding the Equation

The given equation is a quadratic equation in the form of x2+2xβˆ’7=(x+a)2+bx^2 + 2x - 7 = (x + a)^2 + b. To begin solving this equation, we need to expand the right-hand side of the equation. The expanded form of (x+a)2(x + a)^2 is x2+2ax+a2x^2 + 2ax + a^2. Substituting this into the original equation, we get:

x2+2xβˆ’7=x2+2ax+a2+bx^2 + 2x - 7 = x^2 + 2ax + a^2 + b

Simplifying the Equation

Now that we have expanded the right-hand side of the equation, we can simplify it by combining like terms. The x2x^2 terms on both sides of the equation cancel each other out, leaving us with:

2xβˆ’7=2ax+a2+b2x - 7 = 2ax + a^2 + b

Isolating the Terms

Our next step is to isolate the terms containing aa and bb on one side of the equation. To do this, we can subtract 2ax2ax from both sides of the equation and then subtract a2a^2 from both sides. This gives us:

2xβˆ’7βˆ’2ax=a2+b2x - 7 - 2ax = a^2 + b

Rearranging the Terms

We can further simplify the equation by rearranging the terms. Let's move the a2a^2 term to the left-hand side of the equation and the bb term to the right-hand side. This gives us:

2xβˆ’7βˆ’2axβˆ’a2=b2x - 7 - 2ax - a^2 = b

Simplifying the Left-Hand Side

Now that we have rearranged the terms, we can simplify the left-hand side of the equation. We can combine the 2x2x and βˆ’2ax-2ax terms to get:

2xβˆ’2axβˆ’7βˆ’a2=b2x - 2ax - 7 - a^2 = b

Factoring Out the Common Term

We can factor out the common term 2x2x from the first two terms on the left-hand side of the equation. This gives us:

2x(1βˆ’a)βˆ’7βˆ’a2=b2x(1 - a) - 7 - a^2 = b

Simplifying the Equation

Now that we have factored out the common term, we can simplify the equation further. We can combine the βˆ’7-7 and βˆ’a2-a^2 terms to get:

2x(1βˆ’a)βˆ’(7+a2)=b2x(1 - a) - (7 + a^2) = b

Equating the Coefficients

Our goal is to find the values of aa and bb that satisfy the equation. To do this, we can equate the coefficients of the xx terms on both sides of the equation. The coefficient of the xx term on the left-hand side of the equation is 2(1βˆ’a)2(1 - a), and the coefficient of the xx term on the right-hand side of the equation is 00. Equating these coefficients, we get:

2(1βˆ’a)=02(1 - a) = 0

Solving for a

Now that we have equated the coefficients, we can solve for aa. To do this, we can divide both sides of the equation by 22 and then subtract 11 from both sides. This gives us:

1βˆ’a=01 - a = 0

a=1a = 1

Substituting the Value of a

Now that we have found the value of aa, we can substitute it into the original equation. Substituting a=1a = 1 into the equation, we get:

x2+2xβˆ’7=(x+1)2+bx^2 + 2x - 7 = (x + 1)^2 + b

Expanding the Right-Hand Side

We can expand the right-hand side of the equation by squaring the binomial (x+1)(x + 1). This gives us:

x2+2xβˆ’7=x2+2x+1+bx^2 + 2x - 7 = x^2 + 2x + 1 + b

Simplifying the Equation

Now that we have expanded the right-hand side of the equation, we can simplify it by combining like terms. The x2x^2 terms on both sides of the equation cancel each other out, leaving us with:

βˆ’7=1+b-7 = 1 + b

Solving for b

Now that we have simplified the equation, we can solve for bb. To do this, we can subtract 11 from both sides of the equation. This gives us:

βˆ’8=b-8 = b

Conclusion

In this article, we have solved the quadratic equation x2+2xβˆ’7=(x+a)2+bx^2 + 2x - 7 = (x + a)^2 + b to find the values of aa and bb. We found that a=1a = 1 and b=βˆ’8b = -8. This solution satisfies the original equation and provides a clear understanding of the relationship between the coefficients of the quadratic equation.

Final Answer

The final answer is:

Introduction

In our previous article, we solved the quadratic equation x2+2xβˆ’7=(x+a)2+bx^2 + 2x - 7 = (x + a)^2 + b to find the values of aa and bb. In this article, we will answer some frequently asked questions related to the solution of this equation.

Q: What is the significance of the value of a in the quadratic equation?

A: The value of aa in the quadratic equation represents the coefficient of the xx term in the expanded form of (x+a)2(x + a)^2. In the solution of the equation, we found that a=1a = 1. This value of aa allows us to rewrite the equation in a simpler form and solve for the value of bb.

Q: How did you simplify the equation to find the value of b?

A: To simplify the equation, we expanded the right-hand side of the equation by squaring the binomial (x+1)(x + 1). This gave us x2+2x+1+bx^2 + 2x + 1 + b. We then combined like terms and simplified the equation to get βˆ’7=1+b-7 = 1 + b. Finally, we solved for bb by subtracting 11 from both sides of the equation.

Q: What is the relationship between the coefficients of the quadratic equation and the values of a and b?

A: The coefficients of the quadratic equation are related to the values of aa and bb through the expanded form of (x+a)2(x + a)^2. In the solution of the equation, we found that a=1a = 1 and b=βˆ’8b = -8. These values of aa and bb satisfy the original equation and provide a clear understanding of the relationship between the coefficients of the quadratic equation.

Q: Can you provide a step-by-step solution to the quadratic equation?

A: Yes, we can provide a step-by-step solution to the quadratic equation. Here is the solution:

  1. Expand the right-hand side of the equation by squaring the binomial (x+a)(x + a).
  2. Combine like terms and simplify the equation.
  3. Equate the coefficients of the xx terms on both sides of the equation.
  4. Solve for aa by dividing both sides of the equation by 22 and then subtracting 11 from both sides.
  5. Substitute the value of aa into the original equation and expand the right-hand side.
  6. Simplify the equation by combining like terms and solving for bb.

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 correctly.
  • Not combining like terms correctly.
  • Not equating the coefficients of the xx terms on both sides of the equation.
  • Not solving for aa and bb correctly.

Q: Can you provide some examples of quadratic equations that can be solved using this method?

A: Yes, here are some examples of quadratic equations that can be solved using this method:

  • x2+3xβˆ’2=(x+a)2+bx^2 + 3x - 2 = (x + a)^2 + b
  • x2βˆ’4x+3=(x+a)2+bx^2 - 4x + 3 = (x + a)^2 + b
  • x2+2xβˆ’5=(x+a)2+bx^2 + 2x - 5 = (x + a)^2 + b

Conclusion

In this article, we have answered some frequently asked questions related to the solution of the quadratic equation x2+2xβˆ’7=(x+a)2+bx^2 + 2x - 7 = (x + a)^2 + b. We have provided a step-by-step solution to the equation and highlighted some common mistakes to avoid when solving quadratic equations. We have also provided some examples of quadratic equations that can be solved using this method.

Final Answer

The final answer is:

a=1a = 1 b=βˆ’8b = -8