For the polynomial P(x) = x* + 10x – 12x – 20æ +1 apply root squaring method to obtainP (x) :P2(2) =P3(2) =Based on the coefficients of P3(x)The root a1 of largest absolute value for P(x) isThe root ag of second largest absolute value for P(æ) isThe root az of third largest absolute value for P(a) isThe root a4 of smallest absolute value for P(x) is

Answered: For the polynomial P(x) = a* + 10a° –…

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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Question

For the polynomial P(x) = x* + 10x – 12x – 20æ +1 apply root squaring method to obtain
P (x) :
P2(2) =
P3(2) =
Based on the coefficients of P3(x)
The root a1 of largest absolute value for P(x) is
The root ag of second largest absolute value for P(æ) is
The root az of third largest absolute value for P(a) is
The root a4 of smallest absolute value for P(x) is

Expert Solution

Step 1

Given that $P (x) = x^{4} + 10 x^{3} - 12 x^{2} - 20 x + 1$

Using Graeffe's Root squaring Method, then

$P_{1} (x) = x^{4} + 10 x^{3} - 12 x^{2} - 20 x + 1 \Rightarrow x^{4} + 10 x^{3} - 12 x^{2} - 20 x + 1 = 0$

So, put all even powers of x one side of equality sign and all odd powers of x other side of equality sign then

$x^{4} - 12 x^{2} + 1 = - (10 x^{3} - 20 x)$

Take square on both side, we get

${(x^{4} - 12 x^{2} + 1)}^{2} = {(- 10 x (x^{2} - 2))}^{2} \Rightarrow {({(x^{2})}^{2} - 12 x^{2} + 1)}^{2} = 100 x^{2} {(x^{2} - 2)}^{2} put x^{2} = y \Rightarrow {(y^{2} - 12 y + 1)}^{2} = 100 y {(y - 2)}^{2} \Rightarrow y^{4} + 144 y^{2} + 1 - 24 y^{3} - 24 y + 2 y^{2} = 100 y (y^{2} + 4 - 4 y) \Rightarrow y^{4} + 144 y^{2} + 1 - 24 y^{3} - 24 y + 2 y^{2} = 100 y^{3} + 400 y - 400 y^{2} \Rightarrow y^{4} + 144 y^{2} + 1 - 24 y^{3} - 24 y + 2 y^{2} - 100 y^{3} - 400 y + 400 y^{2} = 0 \Rightarrow y^{4} + 124 y^{3} + 546 y^{2} - 424 y + 1 = 0 Hence, P_{2} (x) = x^{4} + 124 x^{3} + 546 x^{2} - 424 x + 1$