Given the roots of the quadratic $2x^2+6x+7=0$ find a quadratic with roots $\alpha^2-1$ and $\beta^2-1$

I was able to solve this problem in two ways:

Method 1:

Sum of the roots $\alpha+\beta=-\frac{b}{a}$

Product of roots $\alpha\beta=\frac{c}{a}$

Hence $\alpha+\beta=-3$ and $\alpha\beta=\frac{7}{2}$

We want an equation with roots $\alpha^2-1$ and $\beta^2-1$

The sum of the roots of the new quadratic will be $\alpha^2-1+\beta^2-1=\alpha^2+\beta^2-2$

The product of the roots of the new quadratic will be $(\alpha^2-1)(\beta^2-1)=\alpha^2\beta^2-(\alpha+\beta)+1$

We are able to compute $\alpha^2+\beta^2$ as it is $(\alpha+\beta)^2-2\alpha\beta$ and so the problem is solved.

Plugging the numbers gives $4u^2+45=0$

Method 2

Let $u=\alpha^2-1\implies\alpha=\sqrt{u+1}$ but we know that $\alpha$ solves the original equation so:


Question: The first method clearly uses the values of $\alpha$ and $\beta$ but the second seemingly only requires $\alpha$. How is this possible? Sure, one man's $\alpha$ is another's $\beta$ and so you could relabel as the choice of $\alpha$ and $\beta $ is arbitrary. This is believable because of the symmetry involved in the new roots $\alpha^2-1$ looks much like a $\beta^2-1$ but I feel there must be more to this. Supposing one root of the new quadratic was $\alpha^2-1$ but the other was $\beta^3-2\beta$ or something worse? How would the second method know? This leads me to a more fundamental question.

Are there only certain functions of the roots of an old quadratic that can we can find a new quadratic in this way? I suppose we could consider $(x-f(\alpha)(x-g(\beta)=0$ where $f$ and $g$ are the functions of the old roots but then could we alway compute these numerically?

Thanks for taking the time to read this and for any contributions.

  • $\begingroup$ The main difference is where you're working. In the first case, you're working in the extension, while you're working in the base field in the second case. $\endgroup$ – Michael Burr May 8 '16 at 10:40
  • $\begingroup$ @michael burr thanks. Going to have to think about your comments. I know what a field extension is and what subfields are but I've never fully grasped Galois correspondence or have a great working knowledge. $\endgroup$ – Karl May 8 '16 at 10:55
  • $\begingroup$ Also, observe that after you pick $\alpha$, you know $\beta$ because $\beta=-\alpha-3$. $\endgroup$ – Michael Burr May 8 '16 at 11:15

Observe the following: Since $\alpha+\beta=-\frac{b}{a}$, you know that $\alpha+\beta=-3$. Above, you considered the case where $u=\alpha^2-1$. Suppose that you have picked a value for $\alpha$ (there are two solutions to the original quadratic and $\alpha$ could be either of them; the Galois action permutes the two roots without changing the base field).

Now, since $\beta=-\alpha-3$, and you know that $\alpha=\sqrt{u+1}$, you know that $\beta=-\alpha-3=-\sqrt{u+1}-3$. Since $2\beta^2+6\beta+7=0$, we know that $2(u+1+6\sqrt{u+1}+9)-6\sqrt{u+1}-18+7=0$. Simplifying, we have that $6\sqrt{u+1}+2u+9=0$. In other words, $$ \sqrt{u+1}=-\frac{2u+9}{6}. $$ This is exactly the same expression that you found above, but starting with $\beta$.

In general, since the Galois action takes $a+b\alpha$ to $a+b\beta$ (where $a$ and $b$ are in $\mathbb{Q}$), we see that the Galois action takes $\alpha^2-1$ to $\beta^2-1$. Therefore, if $\alpha^2-1$ is the root of a polynomial $p(x)$ with coefficients in $\mathbb{Q}$, then so is $\beta^2-1$. More precisely, if $\sigma$ is the Galois action and $p(x)=\sum a_ix^i$, then you know that $p(\alpha)=0$, but $\sigma(p(\alpha))=\sigma(0)=0$, and $\sigma(\sum a_i\alpha^i)=\sum a_i\sigma(\alpha)^i$.

Since $\alpha^2-1$ and $\beta^2-1$ are Galois conjugates, any polynomial (with coefficients in $\mathbb{Q}$) which vanishes at one of them vanishes at both. That is why you only need to know $\alpha$ in the second approach, the $\beta$ comes for free.

  • $\begingroup$ Thanks for your help I will study and then accept. $\endgroup$ – Karl May 8 '16 at 13:18
  • $\begingroup$ Got the jist of this thanks $\endgroup$ – Karl May 8 '16 at 15:49

This is my solution:

Note that $2x^2+6x+7=0$ is equivalent to $x^2= -3x-\dfrac 72$


  1. $\alpha + \beta = -3$

  2. $\alpha \, \beta = \dfrac 72$.

  3. $\alpha^2 - 1 = -3\alpha-\dfrac 92$

  4. $\beta^2 - 1 = -3\beta-\dfrac 92$

We compute

\begin{align} \hline (\alpha^2 - 1) + (\beta^2 - 1) &= (-3\alpha-\dfrac 92)+(-3\beta-\dfrac 92)\\ &= -3(\alpha + \beta) - 9 \\ &= 0 \\ \hline (\alpha^2 - 1)(\beta^2 - 1) &=(-3\alpha-\dfrac 92)(-3\beta-\dfrac 92) \\ &= 9\alpha\beta+\dfrac{27}{2}(\alpha + \beta) + \dfrac{81}{4}\\ &= \dfrac{63}{2} -\dfrac{81}{2}+\dfrac{81}{4} \\ &= \dfrac{45}{4}\\ \hline \end{align}

So the polynomial with roots $\alpha^2-1$ and $\beta^2-1$ is $4x^2 + 45$.

another way to look at your second method

Let $\alpha$ and $\beta$ be the roots of $Ax^2 + Bx + C$.

Let $u = \alpha^2 - 1$

Then $\alpha = \pm \sqrt{u+1}$.

\begin{align} Ax^2 + Bx + C &= 0 \\ A\alpha^2 + B\alpha + C &= 0 \\ A(u+1) \pm B\sqrt{u+1} + C &= 0 \\ \pm B\sqrt{u+1} &= -A(u+1) - C \\ \pm B\sqrt{u+1} &= -Au -(A+C) \\ B^2u + B &= A^2u^2 +2A(A+C)u + (A+C)^2 \\ A^2u^2 +(2A^2-B^2+2AC)u + (A+C)^2 - B &= 0 \end{align}

Letting $\beta = v^2 - 1$ would have resulted in $A^2v^2 +(2A^2-B^2+2AC)v + (A+C)^2 - B = 0$

So it seems that just using $\alpha$ will work.


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