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+(x-1)+(x-1)+qy-2)+ Expanding Equation (3) and simplifying, we obtain the following equation +(c-512)+(a+; За За 2b be 27a2 за + d = 0 = =0 (3)

+(x-1)+(x-1)+qy-2)+ Expanding Equation (3) and simplifying, we obtain the following equation +(c-512)+(a+; За За 2b be 27a2 за + d = 0 = =0 (3)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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 Please Derive (4) from (3) please show step by step and show work

How to Find the Exact Solution of a General Cubic Equation In this chapter, we are going to find the exact solution of a general cubic equation ax³ + bx² + cx+d=0 (1) To find the roots of Equation (1), we first get rid of the quadratic term (x²) by making the substitution x=y= to obtain b 3a where b d{y− b )' + b[{y_b )² +c{y − 3 ) + d. +d=0 Expanding Equation (3) and simplifying, we obtain the following equation - 3 day+[d+ 26³ be 27a² 3a =0 ay³ + C- 26³ bc y² + ² [c-²y+² (d+ 2² -0 = 3a y³ + ey+f=0 Equation (4) is called the depressed cubic since the quadratic term is absent. Having the equation in this form makes it easier to solve for the roots of the cubic equation (Click here to know the history behind solving cubic equations exactly). First, convert the depressed cubic Equation (4) into the form 26³ bc 27a 3a Now, reduce the above equation using Vieta's substitution y=z+= For the time being, the constant s is undefined. Substituting into the depressed cubic Equation (5), we get ( ² + ² ) ² + √² + ² } + s = f= == (₁ + where we assumed. W=z³ y=z+ S= x=y (2) b За (3) =0 (4) Expanding out and multiplying both sides by z³, we get zº +(3s +e)z¹ + f2³ + s(3s+e)2² +s³ = 0 Now, let s= (s is no longer undefined) to simplify the equation into a tri-quadratic equation. 25 + 12³ - 2/7 = 0 (9) By making one more substitution, w=z³, we now have a general quadratic equation which can be solved using the quadratic formula. e³ w² + fw- 27 Once you obtain the solution to this quadratic equation, back substitute using the previous substitutions to obtain the roots to the general cubic equation. wz→y →x (5) (6) (7) (8) (10) (11) (12) Note: You will get two roots for w as Equation (10) is a quadratic equation. Using Equation (11) would then give you three roots for each of the two roots of w, hence giving you six root values for z. But the six root values of z would give you six values of y (Equation (6)); but three values of y will be identical to the other three. So one gets only three values of y, and hence three values of x. (Equation (2))
Transcribed Image Text:How to Find the Exact Solution of a General Cubic Equation In this chapter, we are going to find the exact solution of a general cubic equation ax³ + bx² + cx+d=0 (1) To find the roots of Equation (1), we first get rid of the quadratic term (x²) by making the substitution x=y= to obtain b 3a where b d{y− b )' + b[{y_b )² +c{y − 3 ) + d. +d=0 Expanding Equation (3) and simplifying, we obtain the following equation - 3 day+[d+ 26³ be 27a² 3a =0 ay³ + C- 26³ bc y² + ² [c-²y+² (d+ 2² -0 = 3a y³ + ey+f=0 Equation (4) is called the depressed cubic since the quadratic term is absent. Having the equation in this form makes it easier to solve for the roots of the cubic equation (Click here to know the history behind solving cubic equations exactly). First, convert the depressed cubic Equation (4) into the form 26³ bc 27a 3a Now, reduce the above equation using Vieta's substitution y=z+= For the time being, the constant s is undefined. Substituting into the depressed cubic Equation (5), we get ( ² + ² ) ² + √² + ² } + s = f= == (₁ + where we assumed. W=z³ y=z+ S= x=y (2) b За (3) =0 (4) Expanding out and multiplying both sides by z³, we get zº +(3s +e)z¹ + f2³ + s(3s+e)2² +s³ = 0 Now, let s= (s is no longer undefined) to simplify the equation into a tri-quadratic equation. 25 + 12³ - 2/7 = 0 (9) By making one more substitution, w=z³, we now have a general quadratic equation which can be solved using the quadratic formula. e³ w² + fw- 27 Once you obtain the solution to this quadratic equation, back substitute using the previous substitutions to obtain the roots to the general cubic equation. wz→y →x (5) (6) (7) (8) (10) (11) (12) Note: You will get two roots for w as Equation (10) is a quadratic equation. Using Equation (11) would then give you three roots for each of the two roots of w, hence giving you six root values for z. But the six root values of z would give you six values of y (Equation (6)); but three values of y will be identical to the other three. So one gets only three values of y, and hence three values of x. (Equation (2))
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