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## Integrated math 2

### Course: Integrated math 2 > Unit 3

Lesson 7: Completing the square- Completing the square
- Worked example: Completing the square (intro)
- Completing the square (intro)
- Worked example: Rewriting expressions by completing the square
- Worked example: Rewriting & solving equations by completing the square
- Completing the square (intermediate)
- Solve by completing the square: Integer solutions
- Solve by completing the square: Non-integer solutions
- Solve equations by completing the square
- Worked example: completing the square (leading coefficient ≠ 1)
- Completing the square
- Solving quadratics by completing the square: no solution
- Solving quadratics by completing the square
- Completing the square review

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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 $(x+a{)}^{2}+b$ .

For example, ${x}^{2}+2x+3$ can be rewritten as $(x+1{)}^{2}+2$ . 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.

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

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 ${25}$ .

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

Take the square root of both sides.

Isolate $x$ to find the solution(s).

*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.

First, divide the polynomial by $4$ (the coefficient of the ${x}^{2}$ term).

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 $\frac{5}{2}$ , and squaring it gives us ${{\displaystyle \frac{25}{4}}}$ , our constant term.

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

Take the square root of both sides.

Isolate $x$ to find the solution.

The solution is: $x=-{\displaystyle \frac{5}{2}}$

## Practice

*Want more practice? Check out these exercises:*

## Want to join the conversation?

- 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?(36 votes)- 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.(6 votes)

- I am not understanding how to complete squares when fractions are involved could someone tell me a better way of understanding because the tips or hints given aren't really helping me understand.(8 votes)
- Well, it's pretty much the same as with integers, right? It's just that the arithmetic gets a little hairy. But the process is exactly the same. Keep practicing!(1 vote)

- need to complete the square, the problem is x^2 +10x+blank(5 votes)
- its completing the square .so x^2 + 10 x + 25 - 25 which is equal to( x +5)^2 -25(6 votes)

- 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(6 votes)
- 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(3 votes)
- Notice that the problem reads, "Give the solutions in ascending order."(3 votes)

- Can you just use the quadratic formula for all of these?(2 votes)
- 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.(3 votes)

- 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?(3 votes)
- how to solve x^2-2x-24=0?(2 votes)
- 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.(3 votes)

- 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(2 votes)- 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.(2 votes)

- I am still having trouble with the fractions aspect of this topic. Can some please explain? it would mean the world to me(2 votes)
- (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?(2 votes)