Question:
If the equation $a x^{2}+2 x+a=0$ has two distinct roots, if
(a) $a=\pm 1$
(b) $a=0$
(c) $a=0,1$
(d) $a=-1,0$
Solution:
The given quadric equation is $a x^{2}+2 x+a=0$, and roots are distinct.
Then find the value of a.
Here, $a=a, b=2$ and, $c=a$
As we know that $D=b^{2}-4 a c$
Putting the value of $a=a, b=2$ and, $c=a$
$=(2)^{2}-4 \times a \times a$
$=4-4 a^{2}$
The given equation will have real and distinct roots, if $D>0$
$4-4 a^{2}=0$
$4 a^{2}=4$
$a^{2}=\frac{4}{4}$
$a=\sqrt{1}$
$=\pm 1$
Therefore, the value of $a=\pm 1$
Thus, the correct answer is $(a)$