UPSC NDA Math Model Paper 5

Given equation is f(x)=x2+3|x|+2=0
As we know f(x)=|x|={

−x,	x<0
x,	x≥0.

So given equation becomes,
f(x)=x2+3x+2=0 if x≥0
f(x)=x2−3x+2=0 if x<0
Case .1) x≥0
⇒f(x)=x2+3x+2=0
⇒x=

−3±√32−4(1)(2)

2(1)

⇒x=

−3±1

2

⇒x=

−3+1

2

or x=

−3−1

2

⇒x=−1 or x=−2
Here x≥0 so no real roots are possible
Case.2) x<0
f(x)=x2−3x+2=0
x=

3±√32−4(1)(2)

2(1)

⇒x=

3±1

2

x=

3+1

2

or x=

3−1

2

⇒x=2 or x=1
Here x<0 so no real roots are possible