Need help with an Elementary Math question

Question

If $a+b+c=1$ and $ax^2 + bx + c = 0$ has a unique solution. Find $a,b$, and $c$.

Answer 1

Your question is not clear at all, and not much different from:

If $m = n$ and $p = q$ has a unique solution, find $η$.

You probably want to ask:

If $a,b,c$ are reals and $a+b+c = 1$ and $a x^2 + b x + c = 0$ has a unique solution for $x$, then find $a,b,c$.

In which case you should learn to be more precise next time, and the solution starts as follows:

$a x^2 + b x + c = 0$ has a unique solution for $x$ if and only if either of the following happens:

• $a \ne 0$ and $b^2 - 4ac = 0$. [Why is this so?]
• $a = 0$ and $b \ne 0$. [This corresponds to a non-horizontal straight line.]

In the first case, $0 = b^2 - 4ac = (1-a-c)^2 - 4ac = 1 - 2a - 2c + a^2 - 2ac + c^2$. [Can you continue from here? You can solve for $c$ in terms of $a$.]

In the second case, we also have $b + c = 1$, which together with the derived conditions give all the possible solutions.

Answer 2

Here I suppose $a,b,c\in\mathbb{R}.$
Since $ax^2+bx+c=0$ with $a\not=0$ has a unique solution $$x^2+\left(\frac{b}a\right)x+\left(\frac{c}a\right)=x^2+2tx+t^2$$ for some $t\in\mathbb{R}.$

Now, $4a^2t^2=b^2=4ac$ and we know that $a+b+c=1.$ $$b^2=(a+c-1)^2=4ac$$ $$a^2+c^2+2ac-2a-2c+1=4ac$$ $$a^2+c^2-2ac-2a+2c+1=4c$$ For real solutions of $a, b, c$ we have to assume that $a, c\ge0.$ $$(a-c-1)^2=4c$$ $$a=c+1\pm2\sqrt{c}=(\sqrt c\pm1)^2.$$ Hence we have infinitely many solutions (not a unique solution).

We can easily show that parametric solution $$(a,b,c)\in\{\left(\dfrac{p^2}{(p+q)^2},\dfrac{2pq}{(p+q)^2},\dfrac{q^2}{(p+q)^2}\right):p+q\not=0,\,\ p,q\in\mathbb{R}\}$$ satisfies all the given conditions.