# 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...

• You should put dollar signs around the whole formula, not about the individual symbols, since otherwise you need more effort and the result is badly spaced. Commented Jul 21, 2020 at 16:21
• From the first equation $x^4+2ax^2+2x=0$. Commented Jul 21, 2020 at 16:22
• The Wikipedia article resultant may be helpful for you. Commented Jul 21, 2020 at 16:34

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 $$a=-\frac{17}8$$.

• "Solve the question" is a little short for an answer.
– user65203
Commented Jul 21, 2020 at 17:14

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

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:

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