#the eqs (1) x^2 + ax + b = 0 and (2) x^2 + cx + d = 0 have exactly one common root

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cunning spindleBOT
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jagged hatch
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The calculation itself seems fine (some brackets are needed, though), but I think you also need to somehow use the fact that not only p^2 + ap + b = p^2 + cp + d, but they are also both equal to 0.

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Apart from that, I have another idea.
Suppose f(x) = x^2 + ax + b and g(x) = x^2 + cx + d.
Let's take f(x)g(x). It's a polynomial with exactly one root of multiplicity 2, so itself and its derivative have a common factor. So, maybe that can be done somehow?

vocal mica
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try vieta formula?

tropic pawnBOT
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@neat schooner

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fringe beacon
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consider the roots of the first equation to be t and w, and the roots of the seconds equation to be t and z. (t is the common root)
Then you have t + w = -a, tw = b and t + z = -c, tz = d
Just solve for the second equations other root (in this case z) in terms of the coefficients.

(t + w) - (t + z) = (-a) - (-c)
=> w - z = c - a .... (i)

Now we just gotta represent w in terms of a,b,c or d
tw = b and tz = d
=> t = b/w and t = d/z
=> b/w = d/z => w = bz/d

Sub it in (i)
=> bz/d - z = c - a
=> z(b/d - 1) = c - a
=> z[(b - d)/d] = c - a
=> z = (c - a)d / (b - d)
Option A

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I dont know why yall are complicating it so much tho, derivatives in a basic quadratic equation problem ??

jagged hatch
jagged hatch
neat schooner