set of the positive real roots of the equation x cube -2x square-x+2=0​

2 thoughts on “set of the positive real roots of the equation x cube -2x square-x+2=0​”

1. Answer:

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   Step-by-step explanation:

   The given equation is:

   y=x

   3

   −3x+k

   ⇒y

   ′

   =3x

   2

   −3

   y

   ′

   =0

   ⇒3x

   2

   −3=0

   ⇒x=±1. Thus the stationary points being +1 and −1 they lie outside the given region hence no value of k can make the equation roots lie in the given range of (0,1). ….Answer

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2. Answer:

   …1;

   ….-1

   …2

   there is three roots

   Reply