Completing the Square!

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• June 29th 2010, 01:36 AM
rooney
Completing the Square!
I have a number of completing the square tasks that I need to complete however have no idea where I am heading ha!

Could some one show me a step by step guide on completing the example below and then hopefully I will be able to apply this for the rest of the questions.

4Q² - 28Q + 59

Thank-you
• June 29th 2010, 02:27 AM
sa-ri-ga-ma
$(2Q)^2 - 2\times{2Q}\times{7} + (7)^2 + (...)$

Fill in the bracket.
• June 29th 2010, 02:32 AM
earboth
Quote:

Originally Posted by rooney
I have a number of completing the square tasks that I need to complete however have no idea where I am heading ha!

Could some one show me a step by step guide on completing the example below and then hopefully I will be able to apply this for the rest of the questions.

4Q² - 28Q + 59

Thank-you

I'll show you those step which you can use with nearly every "Completing the square" problems:

$\begin{array}{l}4Q^2-28Q+59=4(Q^2-7Q)+59\\
=4\left(Q^2-7Q+\left(\frac72 \right)^2-\left(\frac72 \right)^2\right)+59\\
=4\left(\left(Q^2-7Q+\left(\frac72 \right)^2\right)-\left(\frac72 \right)^2\right)+59 \\
= 4(Q-\frac72)^2-7+59
= (2Q-7)^2+52\end{array}$

1. Factor out the leading factor of the square from the variables.
2. Divide the leading factor of the linear summand by 2 and suare it.
3. Subtract this new square at once.
4. The first 3 summands in the bracket are a complete square.
5. Expand the bracket and collect the constants.
6. Multiply the complete square by the square-root of the leading factor (why?)
• June 29th 2010, 02:56 AM
sa-ri-ga-ma
The answer should be

$(2Q - 7)^2 + 10$
• June 29th 2010, 02:58 AM
rooney
Ha contradicting answers hey!

I'm trying to work through a procedure i found on the internet so that I can derive the answer myself. Not having a great deal of luck at the mo but i'l get there hopefully
• June 29th 2010, 07:21 AM
earboth
Quote:

Originally Posted by sa-ri-ga-ma
The answer should be

$(2Q - 7)^2 + 10$

You are right! Thanks for spotting my mistake (I forgot to square the 7)

Quote:

Originally Posted by rooney
Ha contradicting answers hey!

...

Here comes the correct calculation:

$\begin{array}{l}
= 4(Q-\frac72)^2-4 \cdot \frac{7^2}{2^2}+59
= (2Q-7)^2+10\end{array}$