I find it frusterating and a bit unfair that in both the review and the videos, the problems they show are equations like 4x^2 +20x + 24, while the problems we are given in the practices include functions: h(x) = x^2 +3x -18. Can someone give me advice on dealing with the functions? I know how they work, but how can I do something to "both sides of the equation" when there is only one side to begin with?

As an expression, you learn to complete the square. As an equation, one of the main purposes of completing the square is to find the roots, so it is where h(x)=0 OR to get an equation in vertex form so that graphing is made easier.

need to complete the square, the problem is x^2 +10x+blank

its completing the square .so x^2 + 10 x + 25 - 25 which is equal to( x +5)^2 -25

In problem 2, the question have the same answer in both X1 and X2. However the problem doesn't automatically consider the answer when it is in X1 = 10, while X2 = 4

Notice that the problem reads, "Give the solutions in ascending order."

Can you just use the quadratic formula for all of these?

You can use the derivative of the equation to find the vertex, by setting the equation equal to 0 and then using this formula: 0=ax^2+bx+c=2ax+b, where x is the x-value of the vertex, and you can plug this in to the original equation to find the y value of the vertex, in essence finding you the vertex of the graph.

how to solve x^2-2x-24=0?

First off you have to factor the trinomial. (x-6)(x+4)=0 I got to that by using the Diamon Method of factoring. But it can be done other ways. One of the other methods can be found here. https://www.khanacademy.org/math/algebra/polynomial-factorization/factoring-quadratics-2/v/factoring-trinomials-by-grouping-4 After I have found the factor I can find the zeros. So what you do is you take each half of the equation and set it equal to zero. x-6=0 x+4=0 After you do this you solve for x. x=6 x=-4 So the zeros of the formula and the partial answer to your question is x=6 and x=-4. I hope this helped you.

I'm having an issue when completing the square with trinomials that all have 'minus' signs. It won't work for me for some reason! Here is what it looks like: x^2-2x-168 Can someone help please

Completing the square just requires you to divide the middle term by 2 and square it, so -2/2 = -1 and (-1)^2 = 1. So you end up with (x^2 -2x + 1) - 168 -1 (you have to subtract 1 to balance the 1 you added). You will end up with (x - 1)^2 - 169.

I am still having trouble with the fractions aspect of this topic. Can some please explain? it would mean the world to me

(by the way, ^2 means squared) x^2 + 2x + 6 = 0 -6 -6 ----------------------- x^2+2x=-6 x^2 + 2x + 2/1(because it's the b of ax squared + bx + c) = -6 + 2/1(you add to both sides) (you do this so you can get a square function) x^2 + 2x + 1 = -5 (x + 1) ^ 2 = -5 you do square root and you get: x + 1 = positive and negative square root of 5 subtract 1 from both sides and you get: x = + and - square root of 5 does this help?

Main content

Algebra 1

Course: Algebra 1 > Unit 14

Lesson 8: More on completing the square

Completing the square review

CCSS.Math:

Completing the square is a technique for factoring quadratics. This article reviews the technique with examples and even lets you practice the technique yourself.

What is completing the square?

Completing the square is a technique for rewriting quadratics in the form left parenthesis, x, plus, a, right parenthesis, squared, plus, b.

For example, x, squared, plus, 2, x, plus, 3 can be rewritten as left parenthesis, x, plus, 1, right parenthesis, squared, plus, 2. The two expressions are totally equivalent, but the second one is nicer to work with in some situations.

Example 1

We're given a quadratic and asked to complete the square.

x, squared, plus, 10, x, plus, 24, equals, 0

We begin by moving the constant term to the right side of the equation.

x, squared, plus, 10, x, equals, minus, 24

We complete the square by taking half of the coefficient of our x term, squaring it, and adding it to both sides of the equation. Since the coefficient of our x term is 10, half of it would be 5, and squaring it gives us start color #11accd, 25, end color #11accd.

x, squared, plus, 10, x, start color #11accd, plus, 25, end color #11accd, equals, minus, 24, start color #11accd, plus, 25, end color #11accd

We can now rewrite the left side of the equation as a squared term.

left parenthesis, x, plus, 5, right parenthesis, squared, equals, 1

Take the square root of both sides.

x, plus, 5, equals, plus minus, 1

Isolate x to find the solution(s).

x, equals, minus, 5, plus minus, 1

Want to learn more about completing the square? Check out this video.

Example 2

We're given a quadratic and asked to complete the square.

4, x, squared, plus, 20, x, plus, 25, equals, 0

First, divide the polynomial by 4 (the coefficient of the x, squared term).

x, squared, plus, 5, x, plus, start fraction, 25, divided by, 4, end fraction, equals, 0

Note that the left side of the equation is already a perfect square trinomial. The coefficient of our x term is 5, half of it is start fraction, 5, divided by, 2, end fraction, and squaring it gives us start color #11accd, start fraction, 25, divided by, 4, end fraction, end color #11accd, our constant term.

Thus, we can rewrite the left side of the equation as a squared term.

left parenthesis, x, plus, start fraction, 5, divided by, 2, end fraction, right parenthesis, squared, equals, 0

Take the square root of both sides.

x, plus, start fraction, 5, divided by, 2, end fraction, equals, 0

Isolate x to find the solution.

The solution is: x, equals, minus, start fraction, 5, divided by, 2, end fraction

Practice

Problem 1

Current

Complete the square to rewrite this expression in the form left parenthesis, x, plus, a, right parenthesis, squared, plus, b.

x, squared, minus, 2, x, plus, 17

Want more practice? Check out these exercises:

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Makayla
Posted 3 years ago. Direct link to Makayla's post “I find it frusterating an...”
I find it frusterating and a bit unfair that in both the review and the videos, the problems they show are equations like 4x^2 +20x + 24, while the problems we are given in the practices include functions: h(x) = x^2 +3x -18.
Can someone give me advice on dealing with the functions? I know how they work, but how can I do something to "both sides of the equation" when there is only one side to begin with?
Button navigates to signup pageComment on Makayla's post “I find it frusterating an...”
(31 votes)
Answer
- David Severin
  Posted 3 years ago. Direct link to David Severin's post “As an expression, you lea...”
  As an expression, you learn to complete the square. As an equation, one of the main purposes of completing the square is to find the roots, so it is where h(x)=0 OR to get an equation in vertex form so that graphing is made easier.
  Comment on David Severin's post “As an expression, you lea...”
  (6 votes)
jepsomad000
Posted 6 years ago. Direct link to jepsomad000's post “need to complete the squa...”
need to complete the square, the problem is x^2 +10x+blank
Button navigates to signup pageComment on jepsomad000's post “need to complete the squa...”
(5 votes)
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- Soham
  Posted 6 years ago. Direct link to Soham's post “its completing the square...”
  its completing the square .so x^2 + 10 x + 25 - 25 which is equal to( x +5)^2 -25
  Button navigates to signup page
  (6 votes)
jansenhuang25
Posted 5 years ago. Direct link to jansenhuang25's post “In problem 2, the questio...”
In problem 2, the question have the same answer in both X1 and X2. However the problem doesn't automatically consider the answer when it is in X1 = 10, while X2 = 4
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Jeff Dodds
  Posted 4 years ago. Direct link to Jeff Dodds's post “Notice that the problem r...”
  Notice that the problem reads, "Give the solutions in ascending order."
  Comment on Jeff Dodds's post “Notice that the problem r...”
  (2 votes)
Clark Fischer
Posted 6 years ago. Direct link to Clark Fischer's post “Can you just use the quad...”
Can you just use the quadratic formula for all of these?
Button navigates to signup pageComment on Clark Fischer's post “Can you just use the quad...”
(2 votes)
Answer
- villanian
  Posted 6 years ago. Direct link to villanian's post “You can use the derivativ...”
  You can use the derivative of the equation to find the vertex, by setting the equation equal to 0 and then using this formula:
  0=ax^2+bx+c=2ax+b, where x is the x-value of the vertex, and you can plug this in to the original equation to find the y value of the vertex, in essence finding you the vertex of the graph.
  Button navigates to signup page
  (3 votes)
Anna W
Posted 5 years ago. Direct link to Anna W's post “What's the difference bet...”
What's the difference between solving a quadratic equation set equal to zero by completing the square and rewriting a quadratic function from standard form to vertex form by completing the square?
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(3 votes)
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eschlichting
Posted 2 months ago. Direct link to eschlichting's post “I learned how to do it a ...”
I learned how to do it a different way. I learned to take the b term and do (b/2) to the second power. Then add that term to both sides and simplify from there. I find this way much easier
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😊
Posted 6 years ago. Direct link to 😊's post “how to solve x^2-2x-24=0?”
how to solve x^2-2x-24=0?
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(2 votes)
Answer
- Coleman M. Vaughn: Hot Topic Studios
  Posted 6 years ago. Direct link to Coleman M. Vaughn: Hot Topic Studios's post “First off you have to fac...”
  First off you have to factor the trinomial.
  (x-6)(x+4)=0
  I got to that by using the Diamon Method of factoring. But it can be done other ways. One of the other methods can be found here. https://www.khanacademy.org/math/algebra/polynomial-factorization/factoring-quadratics-2/v/factoring-trinomials-by-grouping-4
  
  After I have found the factor I can find the zeros.
  So what you do is you take each half of the equation and set it equal to zero.
  
  x-6=0
  x+4=0
  
  After you do this you solve for x.
  
  x=6
  x=-4
  
  So the zeros of the formula and the partial answer to your question is x=6 and x=-4.
  I hope this helped you.
  Comment on Coleman M. Vaughn: Hot Topic Studios's post “First off you have to fac...”
  (3 votes)
Hiba Adil
Posted 5 years ago. Direct link to Hiba Adil's post “I'm having an issue when ...”
I'm having an issue when completing the square with trinomials that all have 'minus' signs. It won't work for me for some reason! Here is what it looks like: x^2-2x-168
Can someone help please
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- David Severin
  Posted 5 years ago. Direct link to David Severin's post “Completing the square jus...”
  Completing the square just requires you to divide the middle term by 2 and square it, so -2/2 = -1 and (-1)^2 = 1. So you end up with (x^2 -2x + 1) - 168 -1 (you have to subtract 1 to balance the 1 you added). You will end up with (x - 1)^2 - 169.
  Button navigates to signup page
  (2 votes)
connor hill
Posted 5 years ago. Direct link to connor hill's post “I am still having trouble...”
I am still having trouble with the fractions aspect of this topic. Can some please explain? it would mean the world to me
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- ???
  Posted 5 years ago. Direct link to ???'s post “(by the way, ^2 means squ...”
  (by the way, ^2 means squared)
  x^2 + 2x + 6 = 0
  -6 -6
  -----------------------
  x^2+2x=-6
  x^2 + 2x + 2/1(because it's the b of ax squared + bx + c) = -6 + 2/1(you add to both sides)
  (you do this so you can get a square function)
  x^2 + 2x + 1 = -5
  (x + 1) ^ 2 = -5
  you do square root and you get: x + 1 = positive and negative square root of 5
  subtract 1 from both sides and you get:
  x = + and - square root of 5
  does this help?
  Comment on ???'s post “(by the way, ^2 means squ...”
  (2 votes)
Cydnie Volton
Posted 4 years ago. Direct link to Cydnie Volton's post “I am so very confused, I ...”
I am so very confused, I am looking up peoples notes online, and trying to figure this out. When completing the square using vertex form. When f(X) is involved I get very confused because I am very used to solving using the constant, when the equation is set to zero. Also my sample equation for this is f(x)=x^2+4x+3. She also used this equation for solving set to zero. I also realized in looking at peoples notes that my class has definitly not gotten to adding a number to a, So something like 2x^2 which I am finding in alot of webpages. I think the reason I am so confused and stuck is due to the "Completing the Square Intermediate" set of questions she assigned, I got a couple f(x) questions and she has not taught us this yet, But wants us all to score very well on these.Also am throwing a monkey wrench in this. When trying to find notes people are getting two different factors which completely defeats the point of this entire process.. I am so confused, I am so sorry to be a bother. I just reaallly don't understand this at all. If anyone could help, Direct me anywhere, I have to have this all done by friday, the assignment here and a workpage, I really hope someone gets to this by then. So if you can help me out that would be amazing, And if you can just direct me to a lesson to look at that will explain this for me, Since I write my own notes and break it down anyway, I will be so greatful. Thank you.
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