Polo's question at Yahoo Answers regarding making a perfect square trinomial

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Hello Polo,

Since the leading coefficient is 5, we may write a perfect square as follows:

$$\left(\sqrt{5}c+k \right)^2=5c^2+2\sqrt{5}kc+k^2$$

Now, we know by equating coefficients, that we require:

$$2\sqrt{5}k=-8\,\therefore\,k=-\frac{4}{\sqrt{5}}\,\therefore\,k^2=\frac{16}{5}$$

Hence:

$$5c^2-8c+\frac{16}{5}=\left(\sqrt{5}c-\frac{4}{\sqrt{5}} \right)^2$$

To Polo and any other guests viewing this topic, I invite and encourage you to post other algebra questions in our http://www.mathhelpboards.com/f2/ forum.

Best Regards,

Mark.
 
Nice , i like it especially because it's different than the 'normal' approach that dictates value of coefficient of x^2 must be 1 to 'complete' a square.

:)