Find $(a,b)$ such that in $x^2+ax+b$, whenever $v$ is a root, then $v^2 - 2$ is also a root

Question

Find $(a,b)$ such that in $x^2+ax+b$, whenever $v$ is a root, then $v^2 - 2$ is also a root $a,b$ are real numbers. Roots may or may not be real. In this question, the aim is to find values of and b ,eg. 2,4.

Answer 1

Is this what you want? If an equation $x^2+ax+b=0$ has roots $v,v^2-2$, then by Vieta's formulas, we have $$v+(v^2-2)=-\frac{a}{1}=-a, v(v^2-2)=\frac{b}{1}=b.$$

Then, we have $(a,b)=(-v^2-v+2,v^3-2v)$.

So, you can get $(a,b)$ if you know the value of $v$.

Edit : OK. The answers for the "new" question is the followings :

Since $f(v)=0\iff v^2=-av-b,$ $$f(v^2-2)=(v^2-2)^2+a(v^2-2)+b=(-av-b-2)^2+a(-av-b-2)+b$$$$=\cdots =v(-a^3+2ab+4a-a^2)+(-a^2b+5b-ab-2a+b^2+4).$$

This has to be zero independently to $v$, so we have $$-a^3+2ab+4a-a^2=0\ \text{and}\ -a^2b+5b-ab-2a+b^2+4=0$$ $$\iff\ a(a^2+a-2b-4)=0\ \text{and}\ ba^2+(b+2)a-(b+1)(b+4)=0$$ $$\iff\ "a=0\ \text{and}\ ba^2+(b+2)a-(b+1)(b+4)=0"\ \text{or}\ "a^2+a-2b-4=0\ \text{and}\ ba^2+(b+2)a-(b+1)(b+4)=0"$$

1) The former leads $(a,b)=(0,-1),(0,-4)$.

2) The latter leads $(a,b)=(-4,4),(-1,-2),(1,-1),(2,1)$. (you can set $b=(a^2+a-4)/2$ in the other equation. Then, you'll get an equation of $a$, so you can get $a$, then set them in $b=(a^2+a-4)/2$.)

Hence, we know that the followings are the necessary conditions : $$(a,b)=(0,-1),(0,-4),(-4,4),(-1,-2),(1,-1),(2,1).$$

On the other hand,

1) The $(0,-1)$ case has $v=1$, so this case is sufficient.

2) The $(0,-4)$ case has $v=-2$, so this case is sufficient.

3) The $(-4,4)$ case has $v=2$, so this case is sufficient.

4) The $(-1,-2)$ case does not have any $v$, so this case is not sufficient.

5) The $(1,-1)$ case has $v=\frac{-1\pm \sqrt 5}{2}$, so this case is sufficient.

6) The $(2,1)$ case has $v=-1$, so this case is sufficient.

Hence, we now reach that the answer is either of the followings :

$(a,b,v)=(0,-1,1),(0,-4,-2),(-4,4,2),(1,-1,\frac{-1\pm\sqrt 5}{2}),(2,1,-1)$.

Note that each of the $(-4,4,2),(2,1,-1)$ cases has an relation of $v=v^2-2$.

Answer 2

From the comments I understand that you are seeking some kind of relationship between $a$ and $b$ not involving $v$.

The polynomial $x^2+ax+b$ has two roots (at most) (real or complex) of the form $x_{1}=\dfrac{-a- \sqrt{a^2-4b}}{2}$ and $x_{2}=\dfrac{-a+\sqrt{a^2-4b}}{2}$. And we are told that, either $x_{1}=x_{2}^2-2$ or $x_{2}=x_{1}^2-2$.

I believe both equations lead to some relation like $$a^3-2a^2-2a-3ab+b^2+5b+4=0.$$

Answer 3

Continuing toufik's solution, one can state the following :

Theorem Let $P=X^2+aX+b$ where $a$ and $b$ are real. Then $P$ factorizes as $(X-v)(X-(v^2-2))$ for some $v\in{\mathbb C}$ if and only if $9-4a \geq 0$, and $b=\frac{3a-5\pm(a-1)\sqrt{9-4a}}{2}$.

Proof of theorem. In one direction, suppose that $P$ factorizes as $(X-v)(X-(v^2-2))$ for some $v\in{\mathbb C}$. Then $X^2+aX+b=(X-v)(X-(v^2-2))$, so

$$ a=-v^2-v+2, \ b=v^3-2v \tag{1} $$

We deduce $(v-1)a=-v^3+3v-2$ and hence $b+(v-1)a=v-2$.

So if $a=1$, we must have $b=(-1)$. Otherwise we can write

$$ v=\frac{a-(b+2)}{a-1} \tag{2} $$

Reinjecting this into (1) we see that

$$ a=-\big(\frac{a-(b+2)}{a-1}\big)^2+\big(\frac{a-(b+2)}{a-1}\big)+2 \tag{3} $$

Expanding and clearing out denominators in (3), we obtain

$$ b^2 + (-3a + 5)b + (a^3 - 2a^2 - 2a + 4)=0 \tag{4} $$

The discriminant is $$(-3a+5)^2-4(a^3 - 2a^2 - 2a + 4)=-4a^3 + 17a^2 - 22a + 9 =(9-4a)((a-1)^2) $$

So that

$$ b=\frac{3a-5\pm(a-1)\sqrt{9-4a}}{2} \tag{5} $$

Note that (5) still holds when $(a,b)=(1,-1)$.

Conversely, suppose that (5) holds, so that $b=\frac{3a-5+\varepsilon(a-1)\sqrt{9-4a}}{2}$ for some $\varepsilon=\pm 1$. Inspired by (2), let us put

$$ v=\frac{-1-\varepsilon\sqrt{9-4a}}{2} \tag{6} $$

Then $v^2=\frac{5-2a-\varepsilon\sqrt{9-4a}}{2}$, so

$$ v^2-2=\frac{1-2a+\varepsilon\sqrt{9-4a}}{2} \tag{7} $$

Multiplying (6) and (7), we see that

$$ v(v^2-2)=\frac{2a-1-(9-4a)+\varepsilon\sqrt{9-4a}(2a-1-1)}{4}= \frac{6a-10+\varepsilon\sqrt{9-4a}(2a-2)}{4}= b \tag{8} $$

and adding (6) and (7), we see that

$$ v+(v^2-2)=-a \tag{9} $$

Then (7) and (8) together imply that $X^2+aX+b=(X-v)(X-(v^2-2))$. This concludes the proof.

Answer 4

Given two numbers $r_1$ and $r_2$, to find a monic quadratic polynomial with $r_1$ and $r_2$ as roots, simply use

$$(x-r_1)(x-r_2)$$

which expands to

$$x^2-(r_1+r_2)x+r_1r_2$$

so that what the problem calls $a$ is $-(r_1+r_2)$, while $b$ is $r_1r_2$.

You just plug in $r_1 = v$ and $r_2=v^2-2$.

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Edit: I interpreted the question so that $v$ was given, and all you needed to do was find $a$ and $b$ from the given $v$. There is another interpretation where it is $a$ and $b$ that are given and you need to find a criterion on $(a,b)$ necessary and sufficient to conclude that (at least one) $c$ exists. That is more complicated, see other good answers.

Answer 5

When $v=x+iy$ then $$a=2-v-v^2=2-x-x^2+y^2-iy(1+2x)$$ is real only if (i) $\ x=-{1\over2}$ or (ii) $\ y=0$, i.e. $v$ is in fact real. In case (i) we have $$b=v(v^2-2)=\left(-{1\over2}+iy\right)\left(-{7\over4}-iy -y^2\right)$$ and therefore $${\rm Im}(b)=y\left(-{5\over4}-y^2\right)\ .$$ Therefore in case (i) the coefficient $b$ is real only if $y=0$, which refers us to case (ii).

We know now that the possible values for $a$ and $b$ are $$a=2-v-v^2,\quad b=v(v^2-2)\qquad(v\in{\mathbb R})\ .\tag{1}$$ We may consider $(1)$ as parametric representation of the admissible pairs $(a,b)$. The following figure shows these pairs for $-2.5\leq v\leq 1.5$.

Answer 6

In the question as now formulated ("whenever $v$ is a root, then $v^2−2$ is also a root"), the quadratic polynomial serves only to define its set $S$ of roots (one or two of them), so one might as well reason for$~S$. It must be closed under the map $x\mapsto x^2-2$; this can be achieved in several ways.

1. $S=\{r\}$ where $r^2-2=r$. Then either $r=-1$ or $r=2$, and since $x+ax+b=(x-r)^2$ in this case, one gets $(a,b)=(2,1)$ respectively $(a,b)=(-4,4)$.

2. $S=\{r,s\}$ where $r^2-2=r=s^2-2$ and $r\neq s$. Now $r$ has the same two possibilities as above, and $s=-r$. Since $x+ax+b=(x-r)(x-s)=x^2-r^2$ in this case, one gets $(a,b)=(0,1)$ respectively $(a,b)=(0,4)$.

3. $S=\{r,s\}$ where $r^2-2=s\neq r=s^2-2$. Now $r$ has to satisfy $r=(r^2-2)^2-2$, which gives the $4$th degree equation $r^4-4r^2-r+1=0$. This looks harder, but we know that the two solutions from 1. will give (unwanted) solutions to this equation, so we can divide $x^4-4x^2-x+1$ without remainder by the polynomial $x^2-x-2$ from 1., giving as quotient the polynomial $x^2+x-1$ for which $r$ has to be a root. Since the same argument holds for $s$, it is clear that $s$ must be the other root, and since $x^2+ax+b=(x-r)(x-s)=x^2+x-1$ in this case one finds a single solution $(a,b)=(1,-1)$ here, even without solving the quadratic equation.

Thus one finds exactly $5$ possible pairs for $(a,b)$, namely $\{(2,1),(-4,4),(0,1),(0,4),(1,-1)\}$.

(Solving the final quadratic in 3., while unnecessary, is not hard: $r=-\frac12-\frac{\sqrt5}2$ and $s=-\frac12+\frac{\sqrt5}2$ is one solution, the negative golden ratio and its conjugate; one can also find the two interchanged.)