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2 years, 9 months ago
2 years, 9 months ago
Hi
|x|= x for x>=0
so we get x^2-7x-18 =0
so we get (x-9)(x+2) =0
but x cannot be less than 0 so x is 9
now |x| =-x for x<0
we get x^2+7x-18=0
we get (x+9)(x-2) =0
we get x as -9 as x=2 >0 which is not possible
so we get 2 solutions
2 years, 9 months ago
@shreyas -9 and 9 will be the only two solutions no, +2 and -2 will be rejected as x<0 and x>0 respectively.
2 years, 9 months ago
In my opinion, it is 4: first take x2 - 7x - 18 and solve , you get x=-2 and 9, then take x2+7x-18 and solve, you get x=+2 and -9. Hence 4. (Mod x means you can take one time +ve and one time -ve as per basic definition of mod)