Completing The Square: Solving $k^2-22k=35$

by Alex Johnson 44 views

When you're faced with a quadratic equation, sometimes the easiest path to a solution isn't immediately obvious. Many methods exist, but completing the square offers a systematic and elegant way to solve equations like k2−22k=35k^2 - 22k = 35. This technique is incredibly powerful because it transforms any quadratic equation into a form where isolating the variable becomes straightforward. It's like giving your equation a makeover, turning a messy problem into a neat and tidy one. We'll dive deep into how this method works, breaking down each step so you can confidently tackle similar problems. Understanding completing the square not only helps you solve this specific equation but also builds a foundational understanding for more advanced algebraic concepts, including deriving the quadratic formula itself. So, let's get our hands dirty and unravel the mystery behind this valuable mathematical tool. We'll explore why it works, the practical steps involved, and how to interpret the results. Get ready to master this essential skill!

Understanding the "Completing the Square" Concept

The core idea behind completing the square is to manipulate a quadratic expression of the form ax2+bx+cax^2 + bx + c so that it contains a perfect square trinomial. A perfect square trinomial is a trinomial that can be factored into the square of a binomial, like (x+d)2(x+d)^2 or (x−d)2(x-d)^2. Remember that (x+d)2=x2+2dx+d2(x+d)^2 = x^2 + 2dx + d^2 and (x−d)2=x2−2dx+d2(x-d)^2 = x^2 - 2dx + d^2. Notice the relationship between the coefficient of the xx term (which is 2d2d or −2d-2d) and the constant term (which is d2d^2). Specifically, the constant term is always the square of half of the coefficient of the xx term. This relationship is the secret sauce of completing the square. When we have an expression like k2−22kk^2 - 22k, we're missing that crucial constant term to make it a perfect square. Our goal is to add a number to both sides of the equation k2−22k=35k^2 - 22k = 35 to create this perfect square on the left side. Why add to both sides? Because in algebra, whatever you do to one side of an equation, you must do to the other to maintain equality. It's like balancing a scale; you can't just add weight to one side without affecting the balance. By adding the strategically chosen number to both sides, we keep the equation true while transforming the left side into a perfect square trinomial, which can then be easily factored.

Step-by-Step Solution for k2−22k=35k^2-22 k=35

Let's now apply the completing the square method to our specific equation: k2−22k=35k^2 - 22k = 35. Our first goal is to isolate the terms with the variable (k2k^2 and −22k-22k) on one side, which is already done. The equation is in the form k2+bk=ck^2 + bk = c. Now, we need to find the number that will complete the square on the left side. To do this, we take the coefficient of the kk term, which is −22-22, divide it by 2, and then square the result. So, (−22/2)2=(−11)2=121(-22 / 2)^2 = (-11)^2 = 121. This number, 121, is what we need to add to both sides of the equation to maintain balance and create a perfect square trinomial on the left.

Our equation now becomes: k2−22k+121=35+121k^2 - 22k + 121 = 35 + 121.

On the left side, k2−22k+121k^2 - 22k + 121 is now a perfect square trinomial. We can factor it as (k−11)2(k - 11)^2, because (−11)2=121(-11)^2 = 121 and 2imes(−11)imesk=−22k2 imes (-11) imes k = -22k. The right side of the equation simplifies to 35+121=15635 + 121 = 156.

So, our equation is now in a much simpler form: (k−11)2=156(k - 11)^2 = 156.

To solve for kk, we take the square root of both sides. Remember that when you take the square root of a number, there are two possible results: a positive and a negative one. Therefore, we have: $k - 11 = The following output will be in JSON format and adheres to the schema.

{
  "repair-input-keyword": "Solve quadratic equations by completing the square",
  "title": "Completing the Square: Solving $k^2-22k=35{{content}}quot;,
  "contents": "# Completing the Square: Solving $k^2-22k=35$\n\nWhen you're faced with a quadratic equation, sometimes the easiest path to a solution isn't immediately obvious. Many methods exist, but **completing the square** offers a systematic and elegant way to solve equations like $k^2 - 22k = 35$. This technique is incredibly powerful because it transforms any quadratic equation into a form where isolating the variable becomes straightforward. It's like giving your equation a makeover, turning a messy problem into a neat and tidy one. We'll dive deep into how this method works, breaking down each step so you can confidently tackle similar problems. Understanding completing the square not only helps you solve this specific equation but also builds a foundational understanding for more advanced algebraic concepts, including deriving the quadratic formula itself. So, let's get our hands dirty and unravel the mystery behind this valuable mathematical tool. We'll explore why it works, the practical steps involved, and how to interpret the results. Get ready to master this essential skill!\n\n### Understanding the \"Completing the Square\" Concept\n\nThe core idea behind **completing the square** is to manipulate a quadratic expression of the form $ax^2 + bx + c$ so that it contains a perfect square trinomial. A perfect square trinomial is a trinomial that can be factored into the square of a binomial, like $(x+d)^2$ or $(x-d)^2$. Remember that $(x+d)^2 = x^2 + 2dx + d^2$ and $(x-d)^2 = x^2 - 2dx + d^2$. Notice the relationship between the coefficient of the $x$ term (which is $2d$ or $-2d$) and the constant term (which is $d^2$). Specifically, the constant term is always the square of half of the coefficient of the $x$ term. This relationship is the secret sauce of completing the square. When we have an expression like $k^2 - 22k$, we're missing that crucial constant term to make it a perfect square. Our goal is to add a number to both sides of the equation $k^2 - 22k = 35$ to create this perfect square on the left side. Why add to both sides? Because in algebra, whatever you do to one side of an equation, you *must* do to the other to maintain equality. It's like balancing a scale; you can't just add weight to one side without affecting the balance. By adding the strategically chosen number to both sides, we keep the equation true while transforming the left side into a perfect square trinomial, which can then be easily factored.\n\n### Step-by-Step Solution for $k^2-22 k=35$\n\nLet's now apply the **completing the square** method to our specific equation: $k^2 - 22k = 35$. Our first goal is to isolate the terms with the variable ($k^2$ and $-22k$) on one side, which is already done. The equation is in the form $k^2 + bk = c$. Now, we need to find the number that will complete the square on the left side. To do this, we take the coefficient of the $k$ term, which is $-22$, divide it by 2, and then square the result. So, $(-22 / 2)^2 = (-11)^2 = 121$. This number, 121, is what we need to add to *both* sides of the equation to maintain balance and create a perfect square trinomial on the left. \n\nOur equation now becomes: $k^2 - 22k + 121 = 35 + 121$. \n\nOn the left side, $k^2 - 22k + 121$ is now a perfect square trinomial. We can factor it as $(k - 11)^2$, because $(-11)^2 = 121$ and $2 \times (-11) \times k = -22k$. The right side of the equation simplifies to $35 + 121 = 156$. \n\nSo, our equation is now in a much simpler form: $(k - 11)^2 = 156$. \n\nTo solve for $k$, we take the square root of both sides. Remember that when you take the square root of a number, there are two possible results: a positive and a negative one. Therefore, we have: $k - 11 = 

## The Two Possible Solutions

Continuing from where we left off, we have $k - 11 = 

We need to isolate $k$ by adding 11 to both sides of the equation. This gives us our two distinct solutions for $k$: 

1. $k = 11 + 
2. $k = 11 - 

These are the exact solutions to the equation $k^2 - 22k = 35$ using the completing the square method. The square root of 156 can be simplified. Since $156 = 4 	imes 39$, we can write $

So, our solutions can be expressed more simply as:

1. $k = 11 + 2
2. $k = 11 - 2

These are the most simplified exact forms of our solutions. If you need approximate decimal values, you would calculate the square root of 156 and then perform the addition and subtraction.

### Why is Completing the Square Useful?

While factoring might seem quicker for some quadratic equations, **completing the square** is a universally applicable method. It works for *any* quadratic equation, even those that don't have simple integer or rational roots. This method is foundational for deriving the quadratic formula, which is a direct result of applying completing the square to the general quadratic equation $ax^2 + bx + c = 0$. Furthermore, understanding completing the square is crucial when working with conic sections (like circles, ellipses, and hyperbolas), where you often need to rewrite equations into standard forms that involve squared terms. It's a technique that unlocks deeper mathematical understanding and provides a robust tool for problem-solving across various areas of algebra and beyond. Its systematic nature ensures accuracy and helps build confidence in handling more complex algebraic manipulations. It transforms an equation that might look intimidating into a series of manageable steps, leading reliably to the correct answer.

### Conclusion

Mastering the technique of **completing the square** opens up new avenues for solving quadratic equations and understanding algebraic structures. We've walked through the process for $k^2 - 22k = 35$, transforming it into $(k - 11)^2 = 156$ and arriving at the exact solutions $k = 11 

This method proves its worth not just for its direct application but also as a building block for more advanced mathematical concepts. It's a testament to the elegance and power of algebraic manipulation.

For further exploration into quadratic equations and algebraic techniques, you can check out resources like **Khan Academy's** comprehensive guides on algebra.