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Problem: Find all values of $b$ for which the equations $1988x^2 + bx + 8891 = 0$ and $8891x^2 + bx + 1988 = 0$ have a common root.

What I have done so far: Let the roots of the first quadratic be $r_1$ and $r_2$, and the roots of the second quadratic be $s_1$ and $s_2$. We have the following equations.

  1. $r_1+r_2 = \frac{-b}{1988}$
  2. $r_1r_2=\frac{8891}{1988}$
  3. $s_1+s_2 = \frac{-b}{8891}$
  4. $s_1s_2 = \frac{1988}{8891}$

Without loss of generality, we can let $r_1=s_1$, but I am not sure how to solve these equations after that. Tips?

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  • $\begingroup$ Oh that looks like the same question, whoops :P $\endgroup$
    – PL Wang
    Jan 2, 2021 at 20:38

3 Answers 3

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hint

Observe that zero is not a root and if $ r\ne 0 $ is a root of the first, $ \frac 1r $ will be a root of the second.

So, we should have $$r=\frac 1r$$

or $ r=\pm1 $.

thus

$$1988\pm b+8891=0$$

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  1. Tip. Euclid's division algorithm (if before the remainder of the last division you have a polynomial of degree $\ge$ 1 then the 2 polynomials share a common root)

  2. Tip. A polynomial of degree n (in the complex field) has exactly n roots (in the complex field). Since the solutions are finite the first polynomial will have 2 roots and the second polynomial will have 2 roots and none of the roots of the first polynomial have to be a root for the second polynomial.

  3. Tip. The solutions for which those 2 polynomials are the same must verify that $x=x^{-1}$, that means that either x=1 or x=-1, in which case b=-10879 or b=10879 (respectively).

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  • $\begingroup$ ... I was first! $\endgroup$
    – yugikaiba
    Jan 2, 2021 at 20:39
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Note that, $$p(x)=1988x^2+bx+8891\implies q(x)=x^2p\left(\frac 1x\right)=8891x^2+bx+1981$$

Since $p(0)\neq 0$ we want $b$ such that $p(x),p\left(\frac 1x\right)$ have a common root.

It is easy to see that this implies that the common root, $r$, satisfies $r=\frac 1r$ so that $r\pm 1$.

And from there, the problem is straight forward.

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