Math 2 question (barrons)

[jumpngo]

<p>In the equation x^2+kx+54=0,
on root is twice the other root.
The value(s) of k is(are)....</p>

<p>Is there a simpler way to solve than...</p>

<p>(x-r)(x-2r) = x^2+(-3r)+2r^2</p>

<p>-3r = k ; r = -k/3
2r^2= 54</p>

<p>then plugging r in.</p>

<p>Thanks</p>

[anhtimmy]

<p>can you give me a specific page number to look at?
I would like to see the question in its entirety.</p>

[Aman_Gulati]

<p>ya of course!</p>

<p>x^2+kx+54=0</p>

<p>Let the roots be a & b.</p>

<p>Since sum of roots=> a+b=(-k)
Product Of Roots=> ab=54</p>

<p>Now since a=2b or b=2a.</p>

<p>3a=-k ---------1
&
2a^2=54
a^2=27
so, a=3^(3/2)</p>

<p>Equating In Equation 1,</p>

<p>k=-3^(5/2)</p>