1 answer

Could anyone out there help me out here: how many positive real roots and how many negative roots does the

Question:

could anyone out there help me out here: how many positive real roots and how many negative roots does the problem have; f(a)=a^5-4a^2-7 (using Descartes rule of signs?

Answers

First count the sign changes of {{{f(a)=a^5-4a^2-7}}} From {{{a^5}}} to {{{-4a^2}}}, there is a sign change from positive to negative From {{{-4a^2}}} to {{{-7}}}, there is no change in sign So there is 1 sign change for the expression {{{f(a)=a^5-4a^2-7}}}. So there is 1 positive real zero ------------------------------------------------ {{{f(-a)=(-a)^5-4(-a)^2-7}}} Now let's replace each {{{a}}} with {{{-a}}} {{{f(-a)=-a^5-4a^2-7}}} Simplify. Note: only the terms with odd exponents will have a sign change. Now let's count the sign changes of {{{f(-a)=-a^5-4a^2-7}}} From {{{-a^5}}} to {{{-4a^2}}}, there is no change in sign From {{{-4a^2}}} to {{{-7}}}, there is no change in sign So there are no sign changes for the expression {{{f(-a)=-a^5-4a^2-7}}} So there are 0 negative real zeros Note: if you graph {{{f(a)=a^5-4a^2-7}}}, you will find that there is indeed one positive real zero..

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