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.

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.
