Find coefficients of polynomials $f(x)=x^2+ax+b$ and $g(x)=x^2+cx+d$ $(a,b,c,d \in \mathbb{R})$

Question

Roots of polynomial $f(x)=x^2+ax+b$ are cubes of the roots of polynomial $g(x)=x^2+cx+d$. Sum and product of roots of polynomial $g(x)$ are equal. Find coefficients $a,b,c,d$ so that polynomial $f(x)$ has double root, and $g(x)$ doesn't have double root.

From the condition that $f(x)$ has double root, discriminant of quadratic equation $x^2+ax+b=0$ must be zero, $$D=a^2-4b=0,b=\frac{a^2}{4}$$ Sum and product of roots of polynomial $g(x)$ are equal, $$x_1^3+x_2^3=x_1^3x_2^3$$ Vieta's formulas for two polynomials, $$x_1+x_2=-a$$ $$x_1x_2=b$$ $$x_1^3+x_2^3=-c$$ $$x_1^3x_2^3=d$$ If we cube first equation, $$(x_1^3+x_2^3)+3x_1x_2(x_1+x_2)=-a^3$$ $$-c-3ba=-a^3$$ From $b=\frac{a^2}{4}$ $$a^3-4c=0,c=\frac{a^3}{4}$$ From $x_1^3+x_2^3=x_1^3x_2^3$ implies $d=-c$ If we cube second equation, $$x_1^3x_2^3=b^3=\frac{a^6}{64}=d=-c$$

I don't know how to find these coefficients.

Thanks for replies.

Answer 1

Let the two roots of $g(x)$ be $p$ and $q$. Since roots of $f(x)$ are cubes of the roots of $g(x)$, $$\begin{align*} p+q &= -c\tag1\\ pq &= d\tag2\\ p^3 + q^3 &= -a\tag3\\ p^3q^3 &= b\tag4 \end{align*}$$ From $(3)$, $$\begin{align*} -a &= p^3 + q^3\\ &= (p+q)(p^2 - pq+q^2)\\ &= (p+q)(p^2 + 2pq + q^2 - 3pq)\\ &= -c[(-c)^2 -3d]\\ a &= c^3 -3cd\tag5 \end{align*}$$ And from $(2)$ and $(4)$, $$b = p^3 q^3 = d^3\tag6$$

Since the sum and the product of roots of $g(x)$ are equal, $$d = -c \tag7$$

Writing $a$ and $b$ in terms of $c$,

• Substitute $(7)$ into $(5)$, $$a = c^3 + 3c^2\tag8$$

• Substitute $(7)$ into $(6)$, $$b = -c^3\tag{9}$$

Since $f(x)$ has double roots, by using $(8)$ and $(9)$, $$\begin{align*} \Delta_f = a^2 - 4b &= 0\\ (c^3 + 3c^2)^2 + 4c^3 &= 0\\ c^4 (c+3)^2 + 4c^3 &= 0\\ c^3 (c^3 + 6c^2 + 9c + 4) &= 0\\ c^3 (c+1)^2(c+ 4) &= 0\\ \end{align*}$$

If $c \in \{0, -4\}$ and using $(7)$, $$\begin{align*} \Delta_g &= c^2 - 4d\\ &= c^2 + 4c\\ &= c(c+4)\\ &= 0 \end{align*}$$ which means $g(x)$ has a double root, contradicting to the given conditions.

Hence $c = -1$. Then $a = 2$, $b=1$ and $d = 1$.

Answer 2

The roots of $f$ are equal, and the coefficients of $f$ are real. So we must have $f(x)=(x-\alpha)^2$ for some real $\alpha$.

The roots of $g$ are all cube roots of $\alpha$, so they must be drawn from the set $\{\sqrt[3]{\alpha},\omega\sqrt[3]{\alpha},\omega^2\sqrt[3]{\alpha}\}$, where $\omega=\frac{-1+i\sqrt{3}}{2}$ is a primitive cube root of unity. Note that $\sqrt[3]{\alpha}$ is the only real number in this set; since the roots of $g$ are distinct, it follows that they cannot all be real. Since $g$ has real coefficients, they must be complex conjugates; thus $g$ has roots $\omega\sqrt[3]{\alpha},\omega^2\sqrt[3]{\alpha}$.

Now, as the sum and product of the roots of $g$ is equal, we have $$(\omega+\omega^2)\sqrt[3]{\alpha}=\omega^3\sqrt[3]{\alpha^2}\\ -\sqrt[3]{\alpha}=\sqrt[3]{\alpha^2}$$ and since $\alpha$ is real this implies that $\alpha^2=-\alpha$, and so $\alpha=-1$. That is, $f(x)=(x+1)^2=x^2+2x+1$, and $g(x)=(x+\omega)(x+\omega^2)=x^2-x+1$.

Answer 3

Hint: Let $\alpha, \beta$ be the roots of $g(x)=0$ then the roots of $f(x)$ are $\alpha^3=\beta^3$ which means that $\alpha^3-\beta^3=0=(\alpha-\beta)(\alpha^2+\alpha\beta+\beta^2)$ and since $\alpha\neq \beta$ you have $\alpha^2+\alpha\beta+\beta^2=0$

Then plug in the condition $\alpha \beta=\alpha+\beta=r$ so that $\alpha^2+\alpha\beta+\beta^2=r^2-r=0$

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Since others have completed the solution, here's my take on a simpler answer using the hint.

You can't have $r=0$, because this would give a double root of $0$, so it must be that $r=1$. Then $g(x)=x^2-x+1$. It is clear that $b=\alpha^3\beta^3=r^3=1$, and then $-a=\alpha^3+\beta^3=(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2)=r(r^2-3r)=-2$ so that $a=2$ so that $f(x)=x^2+2x+1$