The polynomial has $a^2x^2+2(a+1)x+4$ exactly one root. What are the possible values for a?

Question

The polynomial $a^2x^2+2(a+1)x+4$ has exactly one root. What are the possible values for a?

I just want to know how to start.

Answer 1

Completing the square $$a^2x^2+2(a+1)x+4=\left(ax+\frac{a+1}a\right)^2-\left(\frac{a+1}a\right)^2+4$$ $$a^2x^2+2(a+1)x+4=\left(ax+\frac{a+1}a\right)^2+\frac{3a^2-2a-1}{a^ 2}$$ So $3a^2-2a-1=3(a+\frac 13)(a-1)$ must be negative and then $a\in\left\{-\frac{1}{3},1\right\}$ for a single root as Jack D'Aurizio already commented.

Answer 2

The discriminant is zero implies $(a+1)^2-4a^2=(a+1-2a)(a+1+2a)=0$ $a=-1/3,a=1$.

Answer 3

HINT

Recall that $ax^2+bx+c=0$ has exactly one root

$$\iff \Delta=b^2-4ac=0$$