Case 1: x > 0
x^{2} – 7x - 30 = 0
( x - 10 ) ( x - 3 ) = 0
X = 10, x = -3
But x = -3 is not possible as we have considered x > 0, thus 1 solution for this case.
Case 2: x < 0
x^{2} + 7x - 30 = 0
( x + 10 )( x - 3)=0
X = -10 and x = 3
Only x = -10 is permissible.
Thus this equation has 2 real solutions
Alternatively, we can think of the above as a quadratic in |x|
x^{2} – 7|x| - 30 = 0 can be factorized as
(|x| -10) ( |x| + 3) = 0
|x| cannot be -3, |x| can only be 10. X can take 2 real values.
Correct Answer: 2