1
$\begingroup$

The problem: For which values of parameter $a$ the functions $f(x)$ and $g(x)$ have at least $1$ common root.

$f(x)=x^2+ax+1=0$

$g(x)=x^2+x+a=0$

What i did was look at the 2 functions separately and find the roots then equal them and i get the answer.

My question: I am looking for a better way because if the functions were more difficult this method won't be practical.

$\endgroup$
  • $\begingroup$ What was your answer? $\endgroup$ – PM. Apr 14 '17 at 14:56
  • 1
    $\begingroup$ it was $1$ and $-2$ $\endgroup$ – yolo expectz Apr 14 '17 at 14:56
4
$\begingroup$

Suppose $x$ is the common root:

$$ \begin{cases} \begin{align} x^2 + ax + 1 = 0 \\ x^2+x+a=0 \end{align} \end{cases} $$

Subtracting the equations gives:

$$\require{cancel} \cancel{x^2} + ax + 1 - (\cancel{x^2}+x+a) = 0 \quad\iff\quad (x-1)(a-1)=0 $$

Then:

  • either $\,a=1\,$ in which case the equations are in fact identical;

  • or $\,x=1\,$ is the common root which, after substituting back in either equation, gives $a=-2\,$.

$\endgroup$
  • 1
    $\begingroup$ Don't know why I didn't see this. Very nice. +1 $\endgroup$ – user12345 Apr 15 '17 at 1:16
1
$\begingroup$

You didn't say how you obtained your roots but possibly you used the formula for solving quadratic equations. This may illustrate an alternative.

Let the roots of $f$ be $p_1$ and $q_1$. Let the roots of $g$ be $p_2$ and $q_2$.

For $f$ $$p_1+q_1=-a\\p_1q_1=1$$

For $g$ $$p_2+q_2=-1\\p_2q_2=a$$

Case 1:

1 root is common, say $p_1=p_2=p$ and $q_1\neq q_2$ Then $$p+q_1=-a\\pq_1=1$$

and $$p+q_2=-1\\pq_2=a$$

Eliminating $p$ and then $q_1$ leads to $q_2=a$, and also then $q_1=1$ and $p=1$. The above equations then give $a=-2$.

Case 2:

Both roots equal $p_1=p_2=p, q_1=q_2=q$

Then as above $$ p+q=-a \\ p+q=-1 $$ Subtracting gives $a=1$

$\endgroup$
0
$\begingroup$

Well, if their roots are equal, then they're equal at $0$. So, $$x^2+ax+1=0=x^2+x+a\implies ax+1=x+a\implies a(x-1)=x-1\implies a=1$$ This is basically an inspection where we find when the functions are exactly equal (have $2$ imaginary roots in this case). The other answer ($a=-2$) I'm not sure how to easily find.

$\endgroup$
0
$\begingroup$

Plotting $y=-x^2-x$ and $y=-x-\dfrac{1}{x}$, they intersect at $(x,y)=(1,2)$.

That is $a=2$ gives common root $x=1$.

Note the trivial case $a=1$ gives no real roots.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.