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Suppose a quadratic equation $$f(x)=x^2+ax+b$$ $f(x)=0$ has roots as $\alpha$ and $\alpha^2-2$. Then find all possible pairs of $(a,b)$. Here $a,b,\alpha\in\mathbb{R}$

My work:

I found two pairs by equating $\alpha$ with $\alpha^2-2$. Also what I think that suppose there are two roots $c$ and $d$. Then if $c=\alpha$ then $\alpha^2-2$ is coming $d$ and if we put $d=\alpha$ then value of $c$ is coming $\alpha^2-2$. This is what should happen. Otherwise we can just put any real number in $\alpha$ and obtain $\alpha^2-2$ yielding infinite solutions.

Any help is greatly appreciated.

I have a solution but not sure whether it is right or wrong. Pls review.

https://www.sarthaks.com/1236891/the-numbers-of-pairs-a-b-of-real-numbers-such-that-whenever-is-a-root-of-the-equation-x-2-ax-b-0

Vanessa
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  • $$\left[\frac{-a \color{red}{\pm} \sqrt{a^2 - 4b}}{2}\right]^2 - 2 = \left[\frac{-a \color{red}{\mp} \sqrt{a^2 - 4b}}{2}\right].$$ I suggest using the change of variable $$r = \sqrt{a^2 - 4b} \implies r \geq 0.$$ Then, identify all satisfying ordered pairs $(a,r)$. Then, determine which of these $(a,r)$ ordered pairs yield a satisfying ordered pair $(a,b)$. – user2661923 Sep 17 '22 at 04:22
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    For a more interesting variant of such a question see https://math.stackexchange.com/questions/586730/find-a-b-such-that-in-x2axb-whenever-v-is-a-root-then-v2-2-is – Z Ahmed Sep 19 '22 at 05:35
  • For an interesting understanding see:https://math.stackexchange.com/questions/4533740/formation-of-two-families-of-quadratic-equations – Z Ahmed Sep 19 '22 at 05:36
  • @ZAhmed thanks....i got my doubt cleared – Vanessa Sep 19 '22 at 19:03
  • Great! You may give my post an up vote. – Z Ahmed Sep 19 '22 at 19:20

1 Answers1

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You're correct that any real $\alpha$ will result in real $a$ and $b$, but not necessarily all real values of $a$ and $b$ are possible. To check on this, note the roots being $\alpha$ and $\alpha^2 - 2$ means that

$$\begin{equation}\begin{aligned} (x-\alpha)(x-(\alpha^2-2)) & = x^2 + ax + b \\ x^2 + (-\alpha - (\alpha^2-2))x + \alpha(\alpha^2-2) & = x^2 + ax + b \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Thus, by equating the coefficients of the powers of $x$, we get (note this could also have been obtained using the Vieta's formulas),

$$a = -\alpha^2 - \alpha + 2 \; \; \to \; \; \alpha^2 + \alpha + (-2 + a) = 0 \tag{2}\label{eq2A}$$

$$b = \alpha(\alpha^2-2) \tag{3}\label{eq3A}$$

Any real $\alpha$ gives a real value for $b$ in \eqref{eq3A}. However, \eqref{eq2A} is a quadratic in $\alpha$, which means $\alpha$ being a real value requires the discriminant to be non-negative, i.e.,

$$1^2 - 4(-2 + a) \ge 0 \; \; \to \; \; 9 - 4a \ge 0 \; \; \to \; \; a \le \frac{9}{4} \tag{4}\label{eq4A}$$

For each such real $a$ satisfying the above inequality, we then get using the quadratic formula that

$$\alpha = \frac{-1 \pm \sqrt{9 - 4a}}{2} \tag{5}\label{eq5A}$$

For $a = \frac{9}{4}$, this gives one value of $\alpha = -\frac{1}{2}$, while for smaller values of $a$ it gives $2$ values of $\alpha$. Using these value(s) of $\alpha$ in \eqref{eq3A} then gives the corresponding value(s) of $b$.

John Omielan
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  • So how many total pairs are possible$?$ Infinite $???$ – Vanessa Sep 17 '22 at 06:37
  • Can you pls review and explain me this solution https://www.sarthaks.com/1236891/the-numbers-of-pairs-a-b-of-real-numbers-such-that-whenever-is-a-root-of-the-equation-x-2-ax-b-0 – Vanessa Sep 17 '22 at 06:40
  • @EveryonesaysFU Yes, there are infinitely many pairs. As stated, if $a = \frac{9}{4}$, there's just $\alpha = -\frac{1}{2}$, so $b = -\frac{1}{2}\left(\frac{1}{4}-2\right) = \frac{7}{8}$ giving the pair $(a,b) = \left(\frac{9}{4},\frac{7}{8}\right)$. For each real $a \lt \frac{9}{4}$, there are $2$ corresponding values of $b$. – John Omielan Sep 17 '22 at 06:56
  • Can you explain the flaw in the solution whose link i have posted – Vanessa Sep 17 '22 at 06:58
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    @EveryonesaysFU With the other question and solution, there's a fairly subtle, but important, difference between your question and that question. Note your question says the $2$ roots are $\alpha$ and $\alpha^2 - 2$, with no stated relationship between them. However, the other question states that, for some given $a$ and $b$, if either of the roots were called $\alpha$, then $\alpha^2 - 2$ would also be a root. This imposes several quite strict relationships that significantly limits the # of possible values of $(a,b)$, as the answer there indicates with its result of $6$. – John Omielan Sep 17 '22 at 07:17
  • Why isn't there infinite pairs...we can put any alpha and obtain corresponding $\alpha^2-2$ – Vanessa Sep 17 '22 at 07:27
  • @EveryonesaysFU For fixed specific values of $a$ and $b$, if one of the roots is $\alpha$, then in almost all cases (apart from the $6$ mentioned in that other solution), then $\alpha^2 - 2$ is not a root as well. For an example, consider my earlier comment where $\alpha = -\frac{1}{2}$. Then the second root is $\alpha^2 - 2 = \frac{1}{4} - 2 = -\frac{7}{4}$. However, if this second root were called $\alpha$ instead, say $\alpha_1$ to help avoid confusion, then $\alpha_1^2 -2 = \frac{49}{16}-2 = \frac{17}{16}$ is not a root with $(a,b)=\left(\frac{9}{4},\frac{7}{8}\right)$. – John Omielan Sep 17 '22 at 07:35
  • @EveryonesaysFU I hope you now understand the differences, in particular why your question here has infinitely many solutions for $(a,b)$ while the other one has only $6$. However, since you're asking about a different question than what you asked here, if there's anything you're still unsure about, I suggest you ask a new question here specifically about your uncertainties. – John Omielan Sep 17 '22 at 07:39