# Quadratic Question [duplicate]

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?

• Oh that looks like the same question, whoops :P Jan 2, 2021 at 20:38

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

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

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).

• ... I was first! Jan 2, 2021 at 20:39

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.