Quadratic Equations with a common root

Question

$1988x^2 + bx +8991$ and

$8991x^2 +bx + 1988$

have a common root, what are the possible values of b?

I tried getting solutions using the quadratic formula and interchanged the $\pm$ to get the values of b but it seemed looked like i was bashing it and the numbers were getting too high. What is the proper approach to this problem?

Answer 1

We have $$1988x^2+bx+8991-(8991x^2+bx+1988)=0,$$ which gives that the same root it's $1$, which gives $b=-10979$ or $-1$, which gives $b=10979.$

Answer 2

The difference yields $$ 7003(x + 1)(x - 1)=0. $$ Then we have two possibilities. We assume that the characteristic of the coefficient field is different from $47$ and $149$, because otherwise $7003=0$.

Answer 3

We get $8991x^2 +bx + 1988-(1988x^2 + bx +8991)=0$.

$7003x^2-7003=0$. Hence, $x^2-1=0$. We hence have $(x-1)(x+1)=0$, so $x=-1$ or $x=1$. From here, we infer that the common root is $-1$, and that $b=10979$.

Answer 4

$1988x^2 + bx +8991$ and $8991x^2 +bx + 1988$ are reciprocal equations: $\alpha$ is a root of one equation iff $1/\alpha$ is a root of the other equation.

Let $\alpha$ and $\beta$ be the roots of the first equation. Then $1/\alpha$ and $1/\beta$ are the roots of the second equation.

Therefore, if $\alpha$ is a common root of both equations, then $\alpha=1/\alpha$ or $\alpha=1/\beta$. The second possibility cannot occur because $\alpha\beta= 8991 \ne 1$.

Thus, $\alpha=1/\alpha$ and so $\alpha^2=1$, that is, $\alpha = \pm 1$.

Finally, $1988 \pm b +8991=0$ gives $b=\pm (1988+8991)$.