Given, set ={x∈R:(|x|−3)|x+4|=6} As, we already know |x|={
x
‌
x≥0
−x
‌
x<0.
and |x+4|={
x+4
‌
x≥−4
−(x+4)
‌
x<−4.
Case I x<−4 (−x−3)(−x−4)=6⇒(x+3)(x+4)=6 x2+7x+12=6⇒x2+7x+6=0 (x+6)(x+1)=0 x=−6 or x=−1 We will reject x=−1 as, −1>−4 ∴ When x<−4,x=−6 is the solution. Case II −4≤x<0 (−x−3)(x+4)=6⇒−(x+3)(x+4)=6 ⇒‌−(x2+7x+12)=6 ⇒‌x2+7x+18=0 As, the discriminant of this quadratic equation is D=72−4.18=49−72=−23 ∵‌‌D=−23 and D<0 So, no real roots and as per the question, x∈R. No solution when −4≤x<0. Case III x≥0 (|x|−3)|x+4|=6⇒(x−3)(x+4)=6 ⇒‌‌x2+x−12=6 ⇒‌‌x2+x−18=0 x=‌
−1±√1+72
2
=‌
−1±√73
2
We will reject x=‌
−1−√73
2
as ‌
−1−√73
2
<0 and here, x≥0 So, x=‌
−1+√73
2
, when x≥0. ∴x=−6 and x=‌
−1+√73
2
are the two solutions which belong to the set. Hence, number of solutions =2