I'm working through the book Core Maths for Advanced Level on my own, and, after solving the above problem, I'm not getting the same answer as the book.
So, given: $$x^2 + ax + a^2 = 0$$
Using the quadratic formula: $a = 1, b = a, c = a^2$.
Therefore, the discriminant is: $$a^2 – 4(1)(a^2) = a^2 – 4a^2 = -3a^2$$
As we're squaring $a$, it will always be non-negative. However, as we're multiplying by $-3$, the result is always non-positive. To me, that says that if $a = 0$, there are equal real roots, but if $a$ is not equal to $0$, there are no real roots. However, the book only gives the answer as being no real roots.
I'm fairly certain I have this right, especially after substituting zero and non-zero values into the original equation, but I wonder if I'm missing something. Which answer is correct?
Best Answer
I believe you are correct. $D=-3a^2$ so we have the case where $a=0$ with solutions $x_{1}=x_{2}=0$, otherwise non-real solutions due to the discriminant being negative.