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."

In the practice problem x^2-2x+17. I do not understand how you get 16.

To complete the square for "x^2-2x", you need to add 1 "x^2-2x+1" = "(x-1)^2" But, the not change the original polynomial, you then need to subtract 1 from 17 (+ 1 and then -1 = 0, so the polynomial is not changed, we just shifted numbers around). This gets you: "(x-1)^2 + 16" Hope this helps.

how to you solve x squared + 11x + 24

First you move the constant, 24, off to the side, so it looks like x^2 + 11x = 24. Then you halve 11, to get 5.5 Square that, and you get 30.25. Add that to the equation as x^2 + 11x + 30.25 = 24 +30.25. The left part of the equation is now perfect - you would get (x + 5.5)^2 = 54.25. To be entirely honest, I am only partly sure I even got that right. I'm not quite a mathematician, so feel free to correct me. I am also fully aware that this is about a month late.

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.

Main content

Completing the square review

Google Classroom

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...”
  (7 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)
Answer
- 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
(4 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...”
  (3 votes)
Mekenzie.ballard
Posted 6 years ago. Direct link to Mekenzie.ballard's post “In the practice problem x...”
In the practice problem x^2-2x+17. I do not understand how you get 16.
Button navigates to signup pageComment on Mekenzie.ballard's post “In the practice problem x...”
(2 votes)
Answer
- Kim Seidel
  Posted 6 years ago. Direct link to Kim Seidel's post “To complete the square fo...”
  To complete the square for "x^2-2x", you need to add 1
  "x^2-2x+1" = "(x-1)^2"
  But, the not change the original polynomial, you then need to subtract 1 from 17 (+ 1 and then -1 = 0, so the polynomial is not changed, we just shifted numbers around).
  This gets you: "(x-1)^2 + 16"
  Hope this helps.
  Comment on Kim Seidel's post “To complete the square fo...”
  (4 votes)
Enrique Bracamontes
Posted 4 years ago. Direct link to Enrique Bracamontes's post “how to you solve x square...”
how to you solve x squared + 11x + 24
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Allison Lee
  Posted 4 years ago. Direct link to Allison Lee's post “First you move the consta...”
  First you move the constant, 24, off to the side, so it looks like x^2 + 11x = 24. Then you halve 11, to get 5.5 Square that, and you get 30.25. Add that to the equation as x^2 + 11x + 30.25 = 24 +30.25. The left part of the equation is now perfect - you would get (x + 5.5)^2 = 54.25.
  
  To be entirely honest, I am only partly sure I even got that right. I'm not quite a mathematician, so feel free to correct me.
  I am also fully aware that this is about a month late.
  Comment on Allison Lee's post “First you move the consta...”
  (3 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?
Button navigates to signup pageComment on Anna W's post “What's the difference bet...”
(3 votes)
Answer
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
Button navigates to signup pageComment on eschlichting's post “I learned how to do it a ...”
(3 votes)
Answer
😊
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?
Button navigates to signup pageButton navigates to signup page
(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)

Algebra (all content)

Course: Algebra (all content) > Unit 9