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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?
:)
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?
:)