MHB How have I dropped a factor 2 on the square root of 19?

AI Thread Summary
The discussion centers on a mathematical problem involving the equation derived from a geometric context. The equation simplifies to \(30 = 3x^2 - 18x\), leading to the completion of the square method. The solution yields \(x = 3 \pm \sqrt{19}\). A participant expresses confusion about a supposed missing factor of two on the square root of 19, but another clarifies that no factor was actually dropped. The resolution emphasizes that the calculations are correct as presented.
mathlearn
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The main problem is http://mathhelpboards.com/pre-algebra-algebra-2/find-length-dc-19355.html#post88492

In this question

$15 = \dfrac{\left((x+3)+(2x-3)\right)h}{2}=\frac12 ((x+3)+(2x-3))\times((2x-3) -(x+3))=\frac12((2x-3)^2-(x+3)^2)=\frac12(3 x^2-18 x)$

So we get $30=3x^2-18x$

Now using the complete the square method

$x^2-6x=10$
$x^2-6x+9=19$
$x=3\pm\sqrt {19}$

This is what I get , I know that i Have missed a factor of two on the square root of 19, How and why was that dropped?

:)
 
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mathlearn said:
I know that i Have missed a factor of two on the square root of 19
You have not.
 
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