Question

how many roots does (x^n-a)^2 =0 have​

Answer 1

Answer:

∣x∣=x when x>0

∣x∣=−x when x<0

So when x>0:

f(x)=x

2

+3x+2=0

⇒(x+1)(x+2)=0

⇒x=−1,−2

So, no root exists which is >0

When x<0:

f(x)=x

2

−3x+2=0

⇒(x−1)(x−2)=0

⇒x=1,2

So, no root exists which is <0

Thus, no real root exists for the function f(x)=x

2

+3∣x∣+2=0

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