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the procedure for the problem whose cubic equation is given and to find out the real root is there in the interval.
Dear student,
The general cubic equation is
A x3 + B x2 + C x + D = 0
The coefficients A, B, C, D are real or complex numbers with A not 0. Dividing through by A, the equation comes to the form
x3 + b x2 + c x + d = 0
Now we want to reduce the last equation by the substitution
x = y + r
The cubic equation becomes:
(y + r)3 + b (y + r)2 + c (y + r) + d = 0<=> y3 + (3 r + b) y2 + (3 r2 + 2 r b + c) y + r3 + r2 b + r c + d = 0
Now we choose y such that the quadratic term disappears
choose r = -b/3So, with te substitution b x = y - - 3the equation x3 + b x2 + c x + d = 0comes in the form y3 + e y + f = 0
tr vieta-substitution To reduce the last equation we use the Vieta subtitition
1 y = z + s - z
The constant s is an undefined constant for the present. The equation
y3 + e y + f = 0becomes s (z + -)3 + e (z + (s/z)) + f = 0 zexpanding an multiplying through by z3 , we have z6 + (3 s + e) z4 + f z3 + s (3 s + e) z2 + s3 = 0
Now we choose s = -e/3. The equation becomes
z6 + f z3 - e3/27 = 0With z3 = u u2 + f u -e3/27 = 0
This is an easy to solve quadratic equation.
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Sagar Singh
B.Tech, IIT Delhi
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