For what values of the real number \(a\) does the quadratic equation \[x^2 + ax + a = 1\] have distinct real roots?

  1. \(a \neq 2\);

  2. \(a > 2\);

  3. \(a=2\);

  4. all values of \(a\).

We can find out about the roots of this quadratic by considering its discriminant.

Before we do this, we must rearrange the equation into the form “\(ax^2+bx+c=0\)”, so \[x^2 + ax + (a-1) = 0.\] Thus the discriminant is equal to \[a^2 - 4\times 1 \times (a-1) = a^2 - 4a + 4 = (a-2)^2.\]

For the quadratic to have two distinct real roots we need this to be positive and so we need \(a \ne 2\).

The answer is (a).