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viswateja thummala Grade: 11
        

the procedure for the problem whose cubic equation is given and to find out the real root is there in the interval.


 

6 years ago

Answers : (1)

SAGAR SINGH - IIT DELHI
879 Points
										

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

The quadratic term disappears


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/3

So, with te substitution

b
x = y - -
3
the equation

x3 + b x2 + c x + d = 0

comes in the form

y3 + e y + f = 0

Vieta's substitution


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 = 0
becomes

s
(z + -)3 + e (z + (s/z)) + f = 0
z

expanding 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 = 0

With 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




































6 years ago
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