The roots of the equationax^2 + 4x = –2 are real, rational, and equal when a has a value of
(1) 1 (3) 3
(2) 2 (4) 4
If a=1, then $\displaystyle x^2+4x+2=0$ has irrational roots. (Use quadratic formula)
If a=2, then $\displaystyle 2x^2+4x+2=0$. Dividing all terms by 2, we have:
$\displaystyle x^2+2x+1=0$. Now factor to get:
$\displaystyle (x+1)(x+1)=0$. We have 2 equal roots (double root) at x=-1
If you test a=3 and a-4, you'll see that they have imaginary roots.