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WHAT MUST BE ADDED TO 9X^-12X+3 TO MAKE IT A WHOLE SQUARE?

 Sep 28, 2020
 #1
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I will do a similar one for you.

\(4x^2+18x-1\)

 

 

 

\(4x^2+18x-1\\ =4(x^2+\frac{18}{4}x-\frac{1}{4})\\ =4(x^2+\frac{9}{2}x-\frac{1}{4})\\ =4(x^2+\frac{9}{2}x+\left[(\frac{9}{4})^2+\frac{1}{4}\right]-\frac{1}{4})\\ =2^2(x+\frac{9}{4})^2\\ =\left[2(x+\frac{9}{4})\right]^2\\~\\ \text{So to do this I had to add}\\ 4\left[(\frac{9}{4})^2+\frac{1}{4}\right]=4*\frac{81+2}{8}=\frac{81+2}{2}=41.5 \)

 

I could have made a stupid error so if you think something is wrong then let me know.

yours can be done the same way.

 

 

 

LaTex:

4x^2+18x-1\\
=4(x^2+\frac{18}{4}x-\frac{1}{4})\\
=4(x^2+\frac{9}{2}x-\frac{1}{4})\\
=4(x^2+\frac{9}{2}x+\left[(\frac{9}{4})^2+\frac{1}{4}\right]-\frac{1}{4})\\
=2^2(x+\frac{9}{4})^2\\
=\left[2(x+\frac{9}{4})\right]^2\\~\\
\text{So to do this I had to add}\\
4\left[(\frac{9}{4})^2+\frac{1}{4}\right]=4*\frac{81+2}{8}=\frac{81+2}{2}=41.5

 Sep 28, 2020

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