Common root of a cubic and a biquadratic equation

Hint:

The $gcd$ of the two polynomials will also have that common root.

Assume $r$ is the common root. Then,

$$r^3+2ar+2=0tag1$$ $$r^4+2ar^2+1=0tag2$$

Take (1)$cdot$r-(2) to obtain $r=frac12$ and then $a=-frac{17}8$.

This is a much more direct way to solve it but the hint given in the previous answer is much faster:

Let $x^3+2ax+2 = 0$ have roots $x_1, x_2, x_3$ and $x^4+2ax^2+1 = 0$ have roots $x_1, y_1, y_2, y_3$ ($x_1$ is the common root)

Using Vieta's formulas (https://en.wikipedia.org/wiki/Vieta%27s_formulas) which relate the coefficients of a polynomial to its roots:

• i) $x_1+x_2+x_3 = 0$, ii) $x_1+y_1+y_2+y_3 = 0$,

• iii) $x_1x_2x_3 = -2$, iv) $x_1y_1y_2y_3 = 1$

• v) $x_1x_2+x_1x_3+x_2x_3 = 2a$, vi) $x_1y_1+x_1y_2+x_1y_3+y_1y_2+y_1y_3+y_2y_3 = 2a$.

• vii) $ x_1y_1y_2+x_1y_1y_3 + x_1y_2y_3+y_1y_2y_3 = 0$

From multiplying vii) by $x_1$ and iv) we have:

• viii) $x_1^2(y_1y_2+y_1y_3+y_2y_3)+1 = 0$.

From viii) we have:

• ix) $y_1y_2+y_1y_3+y_2y_3 = -1/x_1^2$.

From plugging in ix) into vi) we have:

• x) $2a = x_1(y_1+y_2+y_3)-1/x_1^2$.

Solving for $y_1+y_2+y_3$ using ii) and plugging into x) yields:

$2a = -x_1^2-1/x_1^2 = x_1(x_2+x_3)+x_2x_3 = -x_1^2+x_2x_3$ after using the fact that $x_1+x_2+x_3 = 0$.

so $2a = -x_1^2-1/x_1^2 = -x_1^2+x_2x_3$ and $x_2x_3 = -1/x_1^2$

Which means $x_1x_2x_3 = -1/x_1$

From iii) we have: $x_1x_2x_3 = -1/x_1 = -2$ so $x_1 = 1/2$.

Thus, $2a = -x_1^2-1/x_1^2 = -(1/2)^2-2^2 = -frac{17}{4}$ and $a = boxed{-frac{17}{8}}$

Hint: If $f(x)=0$ and $g(x)=0$ then also $xf(x)-g(x)=0$.