Common root of a cubic and a biquadratic equation The value of $a$ given that the cubic equation
$$x^3+2ax+2=0$$
and the biquadratic equation
$$x^4+2ax^2+1=0$$
have a common root.
I know how to use common root condition for two quadratic equations, But I don't know how to solve this...
 A: Assume $r$ is the common root. Then,
$$r^3+2ar+2=0\tag1$$
$$r^4+2ar^2+1=0\tag2$$
Take (1)$\cdot$r-(2) to obtain $r=\frac12$ and then $a=-\frac{17}8$.
A: Hint: If $f(x)=0$ and $g(x)=0$ then also $xf(x)-g(x)=0$.
A: 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 $a = \boxed{-\frac{17}{8}}$
A: Hint:
The $\gcd$ of the two polynomials will also have that common root.
