Common root of quadratic equations with parameter

Question

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.

Answer 1

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 $a=-2\,$.

Answer 2

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.

Answer 3

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 $a=1$

Answer 4

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.