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Question Number 58045 by ANTARES VY last updated on 17/Apr/19

$${(\mathbf{x}+1)}^4<5{\mathbf{x}}^3+21{\mathbf{x}}^2+17\mathbf{x}+61$$ $$\mathbf{find}\mathbf{the}\mathbf{root}\mathbf{x}?$$

Answered by tanmay last updated on 17/Apr/19

$$x^4+4c_1{x}^3+4c_2{x}^2+4c_{3}x+1<5x^3+21x^2+17x+61$$ $$x^4+4x^3+6x^2+4x+1<5x^3+21x^2+17x+61$$ $$x^4-x^3-15x^2-13x-60<0$$ $$-x^4+x^3+15x^2+13x+60>0$$ $$f\left(x\right)=-x^4+x^3+15x^2+13x+60$$ $$f\left(0\right)=60$$ $$f\left(0\right)>0$$ $$f\left(1\right)>0$$ $$f\left(2\right)>0$$ $$f\left(3\right)>0$$ $$f\left(4\right)>0$$ $$f\left(5\right)=0$$ $$sowhenx\in [0,5)thenf\left(x\right)>0$$

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