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The-number-of-distinct-real-roots-of-equation-x-4-4x-3-12x-2-x-1-0-

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Question Number 31927 by rahul 19 last updated on 16/Mar/18
The number of distinct real roots of equation x^4 −4x^3 +12x^2 +x−1=0.
Thenumberofdistinctrealrootsofequationx4−4x3+12x2+x−1=0.
Answered by MJS last updated on 16/Mar/18
f(x)=x^4 −4x^3 +12x^2 +x−1 f′(x)=4x^3 −12x^2 +24x+1 f′′(x)=12x^2 −24x+24 f′′(x)=0 ⇒ no real solution ⇒ ⇒ f(x) has no inflexion point ⇒ ⇒ it looks like a hanging parabola (because it′s +1×x^4 ) ⇒ ⇒ it has 0, 1 or 2 zeros f′(x) has only 1 zero (because f′′(x)>0 for x∈R), by trying we find −.041<x<−.040 ⇒ ⇒ f(x) has it′s minimum there f(−.045)=−1.02...<0 ⇒ ⇒ f(x) has 2 real zeros (x_1 ≈−.314 x_2 ≈.259)
f(x)=x4−4x3+12x2+x−1f′(x)=4x3−12x2+24x+1f″(x)=12x2−24x+24f″(x)=0⇒norealsolution⇒⇒f(x)hasnoinflexionpoint⇒⇒itlookslikeahangingparabola(becauseit′s+1×x4)⇒⇒ithas0,1or2zerosf′(x)hasonly1zero(becausef″(x)>0forx∈R),bytryingwefind−.041<x<−.040⇒⇒f(x)hasit′sminimumtheref(−.045)=−1.02…<0⇒⇒f(x)has2realzeros(x1≈−.314x2≈.259)
Commented by rahul 19 last updated on 18/Mar/18
thank u sir !
thankusir!

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