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Quick quadratic (Posted on 2020-10-12) Difficulty: 1 of 5
Find the roots of the equation (x-a)(x-b)=(x-c)(x-d), if you know that a+d=b+c=2020 and a≠c.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution | Comment 1 of 5
Dividing by (x-b) and (x-d) gives:

(x-a)/(x-d) = (x-c)/(x-b)

By inspection, this has no solution if a+d=b+c=2020 and a≠c.

So, the only solution is if (x-d) = 0 or (x-b) = 0.

If x-b = 0, then either x-d or x-c =0.  But x-d cannot also = 0, because then a = c

So, b = c = 1010 = x

This gives rise to solutions (a,b,c,d,x) = (a, 1010, 1010, 2020-a, 1010).

Similarly, if x-d = 0, solutions are (1010, b, 2020-b, 1010,1010).

In either case, the only possible x value = 1010

  Posted by Steve Herman on 2020-10-12 10:09:51
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