How to determine no. of real roots of quartic equation??

hi shubhum

no of real roots in quadratic equation  we find out by finding discriminant

D > 0   2 REAL ROOTS

D = 0    1 REAL DISTINCT ROOT

D<0 NO REAL ROOT

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jitender

askiitians expert

Hey jitendra I asked for quartic equation .

For eg. X''4plus 4x''3plus4x''2-2x-56=0

i didnt see quartic equations

yes there is a discriminat for quartic equations also bot formula is too long to remember

and to apply and Descartes sign rule does not tell about the exact number of real roots

but is very useful to when we find the max +ve and max -ve roots so we use it

there is no exact method for finding number of real roots we find it using hit and trial

and graph methods and some time calculus also

for eg. let us take the quartic equation given by you

X^4+ 4x^3+4x^2-2x-56 = f(x)

the number of real roots are 2 one +ve and another -ve

find f(0) = -ve   and f(2) =+ve

so f(x) changes from +ve to - ve  so it cuts the x axis once b/w 0 and 2

there os only one +ve root we can say it by descartes sign rule

now we see for wt value of x : f(x) > 0   by hit and trial method we get f(-4)

so one -ve root b/w 0 and -4   and function is increasing for x<-4

so no roots of f(x) for x < -4

so 2 real roots