Value of $~~a~~$ for which $~~2+\sqrt 3~~$ is a root of $~~f$.

Question

Let us consider a function $~~f(x)=x^3-5x^2+ax-1.~$ Find the value of $~~a~~$ for which $~~2+\sqrt 3~~$ is a root of $~~f$.

My attempt: First way is to put the value $~~x=2+\sqrt 3~~$ in $~~f(x)=0~~$ and evaluate the value of $~~a~~$ which is $~~5.$ But this process is very calculative (Since I am getting only $~~1$ min for each questions in this exam).
Next way is more calculative than this, in which I have consider $~~z,~w~~$ are roots of $~f~$ other than $~~2+\sqrt 3.~~$ Then we have $$z+w+2+\sqrt 3=5~~~\text{ and }~~~zw=\frac{1}{2+\sqrt 3}.$$ Then by solving aforesaid equation, one of $~z~$ or $~w~$ is $~1~$ and $~~f(1)=0~$ gives $~~a=5.$

Now my question, is there any other simple or easier method to solve this? Because the exam in which this question belongs gives a slight less than $~~1$ min for each questions. Please help me to solve this.

Answer 1

You could try to guess that the polynomial has all rational coefficients (i.e. $a\in\mathbb{Q}$). Then apply the irrational root theorem (I think this is what the name is) to conclude that if $2+\sqrt{3}$ is a root, then so is its conjugate, $2-\sqrt{3}$.

Hence we have that $$\begin{cases}z+2-\sqrt{3}+2+\sqrt{3}=5\\z(2+\sqrt{3})(2-\sqrt{3})=1\end{cases}$$ $$\begin{cases} z+4=5\\ z(1)=1\end{cases}$$ Since there does exist a solution, $z=1$, to this system of equations, our assumption was correct.

We can then calculate, using vieta's, that $$a=z(2-\sqrt{3})+z(2+\sqrt{3})+(2-\sqrt{3})(2+\sqrt{3})$$ $$a=4z+1$$ $$a=\boxed{5}$$

Answer 2

Since $2 + \sqrt{3}$ is a root of $x^2 - 4x + 1$, we can consider dividing the original polynomial $f$ by $x^2 - 4x + 1$, and $2 + \sqrt{3}$ will be a root of $f$ if and only if it is a root of the remainder. If we do so, then $x^3 \equiv 4x^2 - x \pmod{x^2 - 4x + 1}$, so $$f \equiv -x^2 + (a-1) x - 1 \pmod{x^2-4x+1}.$$ Now, adding $x^2 - 4x + 1$, we see also $$f \equiv (a-5)x \pmod{x^2-4x+1}.$$ From here, we see that $2 + \sqrt{3}$ is a root of $f$ if and only if it is a root of $(a-5)x$, which is equivalent to $$(a-5) (2+\sqrt{3}) = 0.$$ Since $2 + \sqrt{3} \ne 0$, we see this is further equivalent to $a = 5$.

(So here, the idea is that dividing by $x^2 - 4x + 1$ is simpler than dividing by $x - (2 + \sqrt{3})$ for manual calculations, where the latter is computationally equivalent to evaluating $f(2 + \sqrt{3})$.)

Answer 3

$$f(x)=x^3-5x^2+ax-1$$ $$f(2+\sqrt 3)=0$$

Doing synthetic division,

\begin{array}{r|r|r|r|r} & 1 & -5 & a & -1\\ \sqrt 3 + 2 & 0 & \sqrt 3 + 2 & -\sqrt 3 - 3 & (2a-9)+(a-5)\sqrt 3\\ \hline & 1 & \sqrt 3-3 & -\sqrt 3 +(a-3) & (2a-10)+(a-5)\sqrt 3 \end{array}

$$(2a-10)+(a-5)\sqrt 3 = (a-5)(2+\sqrt 3) = 0 \implies a=5$$

Which means we need to have $a=5$.

Answer 4

If $x^3-5x^2+ax-1=0$, then $x$ can't be zero. So we can divide by it and rearrange, to get $$a=-x^2+5x+\frac{1}{x}$$ Now just put $x=2+\sqrt 3$ to get $a$, using $$-x^2=-7-4\sqrt 3$$ $$5x=10+5\sqrt 3$$ $$\frac{1}{x}=2-\sqrt 3$$ Easy!

Answer 5

Another way that doesn't involve too much computation is to "shift" the polynomial to avoid taking powers of a sum of an integer and an irrational number. Taking $ \ u \ = \ x - 2 \ \rightarrow \ x \ = \ u + 2 \ \ , $ the binomial expansions are $$ (u+2)^3 \ - \ 5·(u+2)^2 \ + \ a·(u+2) \ - \ 1 $$ $$ = \ \ (u^3 \ + \ 3·u^2·2 \ + \ 3·u·2^2 \ + \ 2^3) \ - \ 5·(u^2\ + \ 4u \ + \ 4) \ + \ (au \ + \ 2a) \ - \ 1 $$ $$ = \ \ u^3 \ + \ (6 \ - \ 5)·u^2 \ + \ (12 \ - \ 20 \ + \ a)·u \ + \ (8 \ - \ 20 \ + 2a \ - \ 1 ) $$ $$ = \ \ u^3 \ + \ u^2 \ + \ ( a \ - \ 8)·u \ + \ ( 2a \ - \ 13 ) \ \ . $$ We now want to evaluate this at $ \ u \ = \ \sqrt3 \ $ to find the value(s) of $ \ a \ $ for which $$ (\sqrt3)^3 \ + \ (\sqrt3)^2 \ + \ ( a \ - \ 8)·(\sqrt3) \ + \ ( 2a \ - \ 13 ) \ \ = \ \ 0 $$ $$ \Rightarrow \ \ 3·\sqrt3 \ + \ 3 \ + \ ( a \ - \ 8)·(\sqrt3) \ + \ ( 2a \ - \ 13 ) \ \ = \ \ 0 $$ $$ \Rightarrow \ \ ( a \ - \ 5)·(\sqrt3) \ + \ ( 2a \ - \ 10 ) \ \ = \ \ 0 \ \ . $$ This can be solved consistently by $ \ \mathbf{a = 5} \ \ ; $ as this is a linear equation in $ \ a \ \ , $ the solution is unique. (The last line, understandably, comes out looking like the result from steven gregory's synthetic division.)

[Note that this incidentally shows that $ \ 2 - \sqrt3 \ $ is also a zero of the polynomial, since $ \ u \ = \ -\sqrt3 \ $ produces $ \ ( 5 \ - \ a)·(\sqrt3) \ + \ ( 2a \ - \ 10 ) \ \ = \ \ 0 \ $ and the same solution for $ \ a \ \ . $ Alan Abraham's discovery of the third zero as $ \ x = 1 \ \rightarrow \ u = -1 \ $ yields $$ -1 \ + \ 1 \ + \ ( a \ - \ 8)·(-1) \ + \ ( 2a \ - \ 13 ) \ \ = \ \ 0 \ \ \Rightarrow \ \ a \ - \ 5 \ \ = \ \ 0 \ \ . ] $$