perplexus.info :: Just Math : Common root in quadratic combinations

Common root in quadratic combinations (Posted on 2025-03-08)

Let P(x) = x² - 3x - 7, and let Q(x) and R(x) be two quadratic polynomials also with the coefficient of x² equal to 1. David computes each of the three sums P + Q, P + R, and Q + R and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If Q(0) = 2, then find R(0).

No Solution Yet Submitted by Danish Ahmed Khan

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| Comment 2 of 5 |

Since each pair of P+Q, P+R, and Q+R share different roots, I decided to set P+Q = 2*(x-a)*(x-b), P+R = 2*(x-a)*(a-c), and Q+R = 2*(s-b)*(x-c).

Then ((P+Q)-(P+R))/2 = P makes P(x) = x^2 - 2ax + (ab+ac-bc)

Similarly, Q(x) = x^2 - 2bx + (ab+bc-ac)

And, R(x) = x^2 - 2cx + (ac+bc-ab)

We are given P(x) = x^2 - 3x - 7 and Q(0)=2. Then we must have the following system:

-2a = -3

ab+ac-bc = -7

ab+bc-ac = 2

Solving this yields a=3/2, b=-5/3, and c=-27/19

At this point I stopped, usually these sort of problems have fairly clean answers, but -27/19 feels rather cumbersome compared to 3/2 and -5/3.

Posted by Brian Smith on 2025-03-10 12:36:02

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