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Let $a>2$ be a real number. Solve the equation $$ x^3-2 a x^2+\left(a^2+1\right) x+2-2 a=0 $$

The solution given in the book goes like this:

The trick is to view this as an equation in $a$. The discriminant is $\Delta=4(x-1)^2$, and we get $$ a=\frac{x^2+x}{x} \text { or } a=\frac{x^2-x+2}{x} . $$ These are quadratic equations that can be solved easily.

$\textbf{My Doubt:}$ In the problem, $a$ is a constant and $x$ is the variable, and we are solving for $x$, then how can we view it as an equation of $a$ and not as an equation on $x$?

Ellie_Wong
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  • You can always view this equation as a polynomial equation $f(a, x)=0$ with $f\in \Bbb R[a,x]$. Then $a$ and $x$ are just two unknowns. This is always possible (you just "forget" for a while, that $a$ is a constant - in the end we obtain values for $x$ and $a$ anyway). – Dietrich Burde Jul 02 '23 at 18:36
  • @DietrichBurde what does $f\in \Bbb R[a,x]$ mean? – Ellie_Wong Jul 02 '23 at 18:39
  • It means, that $f$ is a polynomial in $x$ and $a$ with real coefficients. More precisely, we have $f(x,a)=x^3-2ax^2+(a^2+1)x+2-2a$. – Dietrich Burde Jul 02 '23 at 18:40
  • You have, as Dietrich Burde implicitly says, a function $z=f(x,a)$ of two variables and it is asked about the values of $(x,a)$ for which $z=0$. – Piquito Jul 02 '23 at 18:45
  • @DietrichBurde Ok, that's a polynomial in 2 variables, which we can plot in a 3-d graph. But you said to forget about $a$ being a constant and we treat it as variable. Then, if we get some 0s of the polynomial, we will be able to get a relation between $a,x$ (which i think is what given in the solution). Then we again start thinking of $a$ as constant, for which, we will get a constant value for $x$ which is our solution. Is this correct? – Ellie_Wong Jul 02 '23 at 18:46

5 Answers5

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$a$ is an unknown constant greater than $2$ in the formulation you have.

Suppose you are able to find what $a$ has to be in terms of $x$ - that will give you an equation or equations $a=f(x)$, or $f(x)=a$ and you know that if you can solve this equation for $x$ it will provide a solution for the original equation. In effect it separates $a$ and $x$ so they are not entangled, as they are in the original equation.

Of course, in some cases, this strategy will not work, but when it does it can reveal solutions which then enable the original equation to be factored or simplified.

Think about it another way - the strategy of change of variable: to be frivolous, I could try $y=\frac 1{x-a}$ - might lead to a simplification if I were to choose the right substitution. A substitution can mix $x$ and $a$ without causing comment.

Mark Bennet
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  • Ok, so this strategy would work for $n$ variables too right? Say we have an equation where $x$ is the variable we are solving for and we have constants $a_1, a_2, \cdots a_n$. Then we treat $a_i$ as variables and will get a polynomial of $n+1$ variables $f(x,a_1,a_2,\cdots, a_n)=0$ and solving for $0$s we will get a relation between $x$ and $a_i$, and that would be the solution to the actual equation as well, am I correct? – Ellie_Wong Jul 02 '23 at 18:58
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Per your question, it is because the equation in $a$ is a quadratic equation that you already know how to solve by using the quadratic formula while the equation in $x$ is a cubic equation which is much harder to solve. Note that when viewed as an equation in $x$, you can still pull if off by noting that $x = a - 1$ is a zero and use synthetic division to factor it completely.

Wang YeFei
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  • Yes, I know the cubic equation is of course much harder to solve, but what I am asking is that what justifies viewing $a$, which in the actual problem is a constant, as variable that we are solving for, rather than $x$. I hope I'm clearly able to express my question – Ellie_Wong Jul 02 '23 at 18:35
  • Either equation in $a$ or $x$ the relationship is the same, that is what you got in the second box. – Wang YeFei Jul 02 '23 at 18:36
  • Which relation? The relation between $a$ and $x$? – Ellie_Wong Jul 02 '23 at 18:37
  • $a = \dfrac{x^2+x}{x}$ or $a = \dfrac{x^2-x+2}{x}$. – Wang YeFei Jul 02 '23 at 18:38
  • I didn't quite get what you meant by "Either equation in a or x the relationship is the same" Like did you mean, if exchange a in the place of x, it would still be the same equation? – Ellie_Wong Jul 02 '23 at 18:40
  • No, I mean if you solve it as an equation in $x$ or an equation in $a$, you will get the same relationship above between $a$ and $x$. – Wang YeFei Jul 02 '23 at 18:42
  • Ellie Wong, Why do you say, "what justifies viewing a, which in the actual problem is a constant, as variable that we are solving for?" The problem as you stated it just says "solve the equation", but they don't say "for x". – Gordon Jul 02 '23 at 18:47
  • @Gordon They said $a>2$ be a real number which implies $a$ being a constant value, hence it's a equation on $x$ that we are solving for. – Ellie_Wong Jul 02 '23 at 18:49
  • Ellie I see your point. "Let $a>2$ be a real number" seems to suggest that $a$ is a constant. But consider a question phrased as follow. "Let $x>2$ be a real number. Solve the equation $x^2=9", neither of us would blink an eye.

    I think this textbook a poor job in posing the question. They're probably just trying to get you to not be too married to "x is always a variable and a is always a constant."

     – Gordon Jul 02 '23 at 19:03
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While it's true that "a" does not usually represent the variable to be solved for and "x" does not usually represent a constant, the hint is clearly inviting you to do so.

Collecting the coefficients of powers of $a$, you your equation is $$ xa^2 + (-2x^2-2)a + (x^3+x+2) = 0, $$ which highlights that your equation is a quadratic in $a$. If you write down the quadratic formula for the solution, regarding $a$ as the variable, then the discriminant is the term under the square root sign, and it does indeed simplify to the expression given in the hint.

Gordon
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How can we view it as an equation of $a$ and not as an equation on $x$?

Just imagine $x$ is the solution you are looking for. A changing of notation can be useful to make things clearer:

If $x_0$ is a solution for $x$, then $$x_0^3-2 a x_0^2+(a^2+1) x_0+2-2 a=0.$$

Despite the fact that $a$ is a constant, its value was not given in the problem. So, in the above equation (where $x_0$ is supposed to be known), $a$ can be viewed as an unknown. As the equality is satisfied, you can use it in order to write $a$ in terms of $x_0$, namely, $$a=\frac{x_0^2+x_0}{x_0} \text { or } a=\frac{x^2_0-x_0+2}{x_0}$$ as in the proposed solution. Finally, from these equalities, you can get $x_0$ (the solution you want) in terms of $a$.

Pedro
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It may be easier to get your head around this if you don't use the quadratic formula. We are told that $$x^3 - 2 a x^2 + (a^2 + 1) x + 2 - 2 a=0$$ We can factorise this as: $$(-1 + a - x) (-2 + x + a x - x^2)=0$$ We haven't said anything about $a$ or $x$, we just rearranged an expression. Now note that, if the product of two numbers is zero, then one of them must be zero. So either $$-1 + a - x=0$$ or $$-2 + x + a x - x^2=0$$ and I'm sure you can finish up from there. I don't think any of these steps should worry you.

Blitzer
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