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What are the roots of f(x)=(x-6)^2(x+2)^2

User Wbg
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1 Answer

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f(x)=0

⇒(x−6)^2. (x+2) ^2 =0

⇒(x−6)^2=0, (x+2)^2=0

⇒x=6,6 ⇒x=−2,−2.

User Max Elkin
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