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# Completing the square review

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

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

### Example 2

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
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:

## Want to join the conversation?

• 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."
• 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.
• 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.
• 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?
• 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 solve x^2-2x-24=0?
• 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
• 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?