factorise hayden

**johnsy123** · February 21st 2010, 12:00 AM

$4x^2-4\sqrt{5x}+5$

**CaptainBlack** · February 21st 2010, 12:12 AM

Originally Posted by johnsy123

$4x^2-4\sqrt{5x}+5$

Assuming a typo and you mean:

$4x^2-4\sqrt{5}x+5=(2x-\sqrt{5})^2$

CB

**Prove It** · February 21st 2010, 12:23 AM

Originally Posted by johnsy123

$4x^2-4\sqrt{5x}+5$

I assume this is actually

$4x^2 - 4\sqrt{5}x + 5$ .

Remember that $(a + b)^2 = a^2 + 2ab + b^2$ .

What you have posted is a perfect square, because

$4x^2 - 4\sqrt{5}x + 5 = (2x)^2 + 2(2x)(-\sqrt{5}) + (-\sqrt{5})^2$

$= (2x - \sqrt{5})^2$

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