 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  As easy as abc (Posted on 2011-06-28) Given an equation ax^2+bx+c=0 what values a,b cause the roots of the equation to be a and b and the discriminant to be equal to c?

 No Solution Yet Submitted by Ady TZIDON Rating: 5.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Rooting for trouble | Comment 3 of 10 | In case anyone was wondering, it doesn't work out nicely into real numbers. Note if a=0 or b=0, then c=-a^2 b implies c=0. But then c=b^2-4ac gives b=0, which means a=0 since the roots of ax^2=0 are zero.

Solving the equation c=b^2-4ac for c gives (4a+1)c=b^2; plugging in c=a^2 b gives (4a+1)(a^2 b)=b^2, so (4a+1)(a^2)=b. Plugging in b in terms of a gives (4a+1)(a^2)=-a^2/(a+1) and so (4a+1)(a+1)=-1 and finally 4a^2+5a+1=-1

At this point, one can notice this equation has no real roots, and thus a and b will be complex. I did some extra computations to check the work, giving
a= -0.625 + 0.330718914 i
b=0.125 + 0.992156742 i
c=0.4453125 + 0.227369253 i

For the other set of values, simply change the plus to minus.

 Posted by Gamer on 2011-06-30 00:09:09 Please log in:

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