Proving the existence of a square root of $ -1_{A} $ in a $ 2 $-dimensional unital algebra $ A $ over $ \Bbb{R} $. Suppose that $ A $ is a $ 2 $-dimensional unital algebra over $ \Bbb{R} $ with a basis $ \{ 1_{A},u \} $, and assume that $ A $ does not have any zero divisors. Show that $ A $ contains an element $ b $ such that $ b^{2} = - 1_{A} $.

I’m sure this is straightforward. I’ve tried thinking about it in terms of a bijective map to show the existence of the element and by writing
$$
b = \lambda_{1} 1_{A} + \lambda_{2} u, \qquad \lambda_{1},\lambda_{2} \in \Bbb{R},
$$
and then evaluating $ b^{2} $, but I didn’t get anywhere useful.
Any help appreciated.
 A: Here $a,b,c,d$ will be real constants.
Suppose $u^2=c+du$. Suppose also that $d=0$. If $c<0$, then we are done. If $c\geq 0$, then $(u+\sqrt{c})(u-\sqrt{c})=0$, contradicting the assumption that there are no zero divisors. Thus we must have that $d\neq 0$.
Now
$$(a+bu)^2=a^2+b^2c+(2ab+b^2d)u$$
Take $a=-\frac{1}{2}bd$, so that
$$(a+bu)^2=a^2+b^2c$$
If $a^2+b^2c\geq 0$, then again we have a zero divisor. Thus we must have that $a^2+b^2c<0$, so $a+bu$ is proportional via a real constant to an element we desire, hence we are done.
A: Not sure how old Frobenio handled this one, but here's how I do it:
Since $A = \text{span}(\{ 1_A, u \})$, we have
$u^2 = \alpha 1_A + \beta u, \;\; \alpha, \beta \in \Bbb R. \tag{1}$
Then
$u^2 - \beta u =  \alpha 1_A, \tag{2}$
or, "completing the square", 
$u^2 - \beta u + \dfrac{\beta^2}{4} 1_A = \alpha 1_A + \dfrac{\beta^2}{4} 1_A = (\alpha + \dfrac{\beta^2}{4})1_A, \tag{3}$
or
$(u - \dfrac{\beta}{2}1_A)^2 =  (\alpha + \dfrac{\beta^2}{4})1_A. \tag{4}$
If
$\alpha + \dfrac{\beta^2}{4} \ge 0, \tag{5}$
there exists $\gamma \in \Bbb R$ such that
$\gamma^2 = \alpha + \dfrac{\beta^2}{4}, \tag{6}$
whence (4) becomes
$(u - \dfrac{\beta}{2}1_A)^2 =  \gamma^2 1_A \tag{7}$
or
$(u - \dfrac{\beta}{2}1_A)^2 - \gamma^2 1_A = 0 \tag{8}$
or
$(u -\dfrac{\beta}{2} 1_A - \gamma 1_A)(u - \dfrac{\beta}{2}1_A + \gamma 1_A) = 0 \tag{9}$
or
$(u -(\dfrac{\beta}{2}  + \gamma)1_A)(u - (\dfrac{\beta}{2} - \gamma) 1_A) = 0; \tag{10}$
neither factor in (10) is $0$ by the linear independence of the basis $\{1_A, u \}$; thus each is a zero divisor in $A$.  If we rule out the existence of zero divisors in $A$, then we must have
$\alpha + \dfrac{\beta^2}{4} < 0; \tag{11}$
in this case we may find $\gamma \in \Bbb R$ such that
$-\gamma^2 = \alpha + \dfrac{\beta^2}{4} \tag{12}$
whence
$(u - \dfrac{\beta}{2} 1_A)^2 = -\gamma^2; \tag{13}$
or 
$(\dfrac{u - \dfrac{\beta}{2}1_A}{\gamma})^2 = \dfrac{(u - \dfrac{\beta}{2} 1_A)^2}{\gamma^2} = -1_A, \tag{14}$
as sought.  QED.
A: For some $a,c\in\mathbb{R}$, you know that $$u^2=a\cdot1+c\cdot u$$ 
So that tells you that $$u^2-c\cdot u-a\cdot1=0$$ You have been told that the algebra has no zero-divisors, so viewed as a polynomial in a variable $u$, this expression does not factor over $\mathbb{R}$. That tells you that $c^2+4a$ is negative. As it will be needed later, this means $c^2/4+a$ is negative.
Now consider $s,t\in\mathbb{R}$, and what $(s\cdot1+t\cdot u)^2$ looks like:
$$\begin{align}
(s\cdot1+t\cdot u)^2&=s^2\cdot1+2st\cdot u+t^2a\cdot1+t^2c\cdot u\\
&=(s^2+t^2a)\cdot1+t(2s+tc)\cdot u
\end{align}$$
We can reexamine this equation in the specific case that $s=-tc/2$, still leaving $t$ free. That leaves 
$$\begin{align}
(s\cdot1+t\cdot u)^2&=(s^2+t^2a)\cdot1\\
&=t^2\left(c^2/4+a\right)\cdot1
\end{align}$$
We already established $c^2+4a$ is negative. You can now choose $t$ appropriately to rescale $t^2\left(c^2/4+a\right)$ to be $-1$.
All this means that you can let $b=s\cdot1+t\cdot u$, and you have the opportunity to set $s$ and $t$ appropriately to cause $b^2=-1\cdot1$. Remembering that $a$ and $c$ are structure constants of the algebra, not up to you, the value for $b$ is $\frac{-c}{2\sqrt{-(c^2/4+a)}}\cdot 1+\frac{1}{\sqrt{-(c^2/4+a)}}\cdot u$.
A: Since $A$ has no zero divisor, multiplication by any nonzero element is a linear isomorphism, in particular $A$ is a division algebra. Since it is generated as an algebra by one element it is also commutative. So $A$ is a field.
Let $b\in A\notin \mathbb{R}\cdot 1_A$. Then $b$ is algebraic over $\mathbb{R}$, which means its minimal polynomial has degree one or two. It can't have degree one, so it must be of degree two, with negative discriminant. So
$$
b^2+2pb+q\cdot1_A=0
$$
with $p^2-q<0$, for some $p,q\in\mathbb{R}$. Then
$$
b^2+2pb+p^2\cdot1_A=(p^2-q)\cdot1_A
$$
and so
$$
\left(\frac{1}{\sqrt{q-p^2}}(b+p\cdot1_A)\right)^2=-1_A
$$
