# Math Help - Help me with Algebra 2. (completing the square)

1. ## Help me with Algebra 2. (completing the square)

Hi. I need help for an upcoming test tomorrow. I did some practice problems, can you guys please check if they're right? And if not, can you please do it and show me your work? I feel like I get the basic concept but I'm missing a few steps.

Complete the square for each expression. Write the resulting expression as a binomial squared.

23. $x^2 - 18x + _$

My answer $(x-9)^2$

24. x^2 + 10x + _

My answer $(x+5)^2$

25. $x^2 - 1/2x + _$

My answer $(x+1/4)^2$

Solve each equation by completing the square (I feel I got these questions wrong)

26. $x^2 + 2x = 7$

My answer $x=5 x=-7$

27. $x^2 - 4x = -1$

My answer $x=4 x=0$

I will also post questions about writing each function in vertex form, but right now I'm focused on this. Can you guys please see if I got anything wrong and if I did, can you please show me how you got the right answer? Also, it would be nice if you guys could explain it or even show a video. This is a great community and I would like to thank everybody who contributes on this website!

2. ## Re: Help me with Algebra 2. (completing the square)

Originally Posted by WannaBSmart
Hi. I need help for an upcoming test tomorrow. I did some practice problems, can you guys please check if they're right? And if not, can you please do it and show me your work? I feel like I get the basic concept but I'm missing a few steps.

Complete the square for each expression. Write the resulting expression as a binomial squared.

23. $x^2 - 18x + _$

My answer $(x-9)^2$

24. x^2 + 10x + _

My answer $(x+5)^2$

25. $x^2 - 1/2x + _$

My answer $(x+1/4)^2$

Solve each equation by completing the square (I feel I got these questions wrong)

26. $x^2 + 2x = 7$

My answer $x=5 x=-7$

27. $x^2 - 4x = -1$

My answer $x=4 x=0$

I will also post questions about writing each function in vertex form, but right now I'm focused on this. Can you guys please see if I got anything wrong and if I did, can you please show me how you got the right answer? Also, it would be nice if you guys could explain it or even show a video. This is a great community and I would like to thank everybody who contributes on this website!
Well your answers are obviously incorrect, seeing as you haven't got the constant on the end of them.

For the first for example, \displaystyle \begin{align*} ( x - 9 )^2 = x^2 - 18x + 81\end{align*}, not \displaystyle \begin{align*} x^2 - 18x \end{align*}.

If you wanted to complete the square, then you would need

\displaystyle \begin{align*} x^2 - 18x = x^2 - 18x + (-9)^2 - (-9)^2 &= (x - 9)^2 - 81 \end{align*}.

Try to fix the rest.

3. ## Re: Help me with Algebra 2. (completing the square)

To complete the square you need to find the correct constant term to give the form $(x+a)^2$

To do this, take the coefficient of $x$ , half it then square it.

4. ## Re: Help me with Algebra 2. (completing the square)

Thank you "Prove It." I will rework the problems and try them again. If I'm getting this right, the only thing I'm doing wrong is not adding the constant at the end? So the answers would be:

23. (x-9)^2 -81
24. (x+5)^2 +25
25. (x+1/4)^2 + 1/6

5. ## Re: Help me with Algebra 2. (completing the square)

Originally Posted by pickslides
To complete the square you need to find the correct constant term to give the form $(x+a)^2$

To do this, take the coefficient of $x$ , half it then square it.
I realize this, but the problem is meant for me to find x. Or am I missing something?

6. ## Re: Help me with Algebra 2. (completing the square)

Originally Posted by WannaBSmart
Thank you "Prove It." I will rework the problems and try them again. If I'm getting this right, the only thing I'm doing wrong is not adding the constant at the end? So the answers would be:

23. (x-9)^2 -81
24. (x+5)^2 +25
25. (x+1/4)^2 + 1/6
I re-read your post "prove it" and then looked at the problem again. I guess I just misread it. I didn't have to factor it all I had to do was do what pickslides did. ;p

7. ## Re: Help me with Algebra 2. (completing the square)

Any help with questions 26 and 27?

8. ## Re: Help me with Algebra 2. (completing the square)

Originally Posted by WannaBSmart
Any help with questions 26 and 27?
Here's a start on #26,

$x^2+2x=7$

$x^2+2x+1-1=7$

$x^2+2x+1=8$

$(x+1)^2=8$

then take the square root of both sides and take 1.

Now apply the same thinking to #27.

9. ## Re: Help me with Algebra 2. (completing the square)

Originally Posted by WannaBSmart
Thank you "Prove It." I will rework the problems and try them again. If I'm getting this right, the only thing I'm doing wrong is not adding the constant at the end? So the answers would be:

23. (x-9)^2 -81
24. (x+5)^2 +25
25. (x+1/4)^2 + 1/6
You're supposed to subtract the constant, not add it. Try expanding your results and collecting like terms to check your answer.

I don't think you understand what is happening when you complete the square. You are adding a CLEVERLY DISGUISED ZERO, since adding zero does not change its value.

Anyway, looking at the general case for a monic quadratic

\displaystyle \begin{align*} (x + n)^2 &= x^2 + 2n\,x + n^2 \end{align*}

Notice that the coefficient of \displaystyle \begin{align*} x \end{align*} and the constant have something in common, \displaystyle \begin{align*} n \end{align*}.

So that means if you only had the first two terms of the quadratic, you could use the second term to find the constant.

Notice that to turn \displaystyle \begin{align*} 2n \end{align*} into \displaystyle \begin{align*} n^2 \end{align*}, you need to HALVE it, then SQUARE it.

But remember that you are planning to add ZERO, so whatever you do add, you need to then subtract.

So let's look at an example. \displaystyle \begin{align*} x^2 + 6x \end{align*}

We need to halve and then square the coefficient of \displaystyle \begin{align*} x \end{align*}. This will be what we add and subtract.

\displaystyle \begin{align*} x^2 + 6x &= x^2 + 6x + 3^2 - 3^2 \\ &= (x + 3)^2 - 9 \end{align*}

Another example: \displaystyle \begin{align*} x^2 + 9x + \frac{5}{4} \end{align*}

We need to halve and square the coefficient of \displaystyle \begin{align*} x \end{align*}. This will be what we add and subtract.

\displaystyle \begin{align*} x^2 + 9x + \frac{5}{4} &= x^2 + 9x + \left(\frac{9}{2}\right)^2 - \left(\frac{9}{2}\right)^2 + \frac{5}{4} \\ &= \left(x + \frac{9}{2}\right)^2 - \frac{81}{4} + \frac{5}{4} \\ &= \left(x + \frac{9}{2}\right)^2 - 19 \end{align*}

Another example: \displaystyle \begin{align*} x^2 - 8x + 2 \end{align*}. Notice that the coefficient of \displaystyle \begin{align*} x \end{align*} is negative. This doesn't change the process, but we do need to remember that the negative is there. Anyway, adding and subtracting half of the coefficient of \displaystyle \begin{align*} x \end{align*} gives

\displaystyle \begin{align*} x^2 - 8x + 2 &= x^2 - 8x + (-4)^2 - (-4)^2 + 2 \\ &= \left[x + (-4)\right]^2 - 16 + 2 \\ &= (x - 4)^2 - 14 \end{align*}

And a final example, I hope you've noticed that completing the square only works for monic quadratics (i.e. where the coefficient of \displaystyle \begin{align*} x^2\end{align*} is 1). What happens if it's something else? You need to take it out as a common factor first. So using \displaystyle \begin{align*} 2x^2 + 3x + \frac{1}{8}\end{align*} as an example...

\displaystyle \begin{align*} 2x^2 + 3x + \frac{1}{8} &= 2\left(x^2 + \frac{3}{2} + \frac{1}{16}\right) \\ &= 2\left[x^2 + \frac{3}{2} + \left(\frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2 + \frac{1}{16}\right] \\ &= 2\left[\left(x + \frac{3}{4}\right)^2 - \frac{9}{16} + \frac{1}{16}\right] \\ &= 2\left[\left(x + \frac{3}{4}\right)^2 - \frac{1}{2}\right] \\ &= 2\left(x + \frac{3}{4}\right)^2 - 1\end{align*}

So now you need to fix your answers.