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and = 4 p(p- 1)x² (x² + y+²)*² +2 p(x² + y° +2)P!, -= 4 p(p-1) y (x + ỷ +z')*² +2 p(x² + yỷ + ?)®, = 4 p(p-1) z² (x² + y² + z²)? +2 p(x° + y° + *). Thus, V²ø = 0 gives 4p(p-1)x (x + у+7)2 +2 p(x? + ў +7) +4 p(p-1) y (x + y + 2)*² +2 p(x² + y + ?)*1 +4 p(p-1)z (x + y + 2)"?+2 p(x + ỷ + р-2 = 0 Which can simplified to [4 p(p-1) + 6 p](x² + y} + z)* = = 0. The above equation is true for x, y and z (except possibly at (0,0,0)) if the constant p is selected such that 4 p(p-1) + 6 p= 0= p(4 p+ 2) = 0= p=0, p=-1/2.

and = 4 p(p- 1)x² (x² + y+²)*² +2 p(x² + y° +2)P!, -= 4 p(p-1) y (x + ỷ +z')*² +2 p(x² + yỷ + ?)®, = 4 p(p-1) z² (x² + y² + z²)? +2 p(x° + y° + *). Thus, V²ø = 0 gives 4p(p-1)x (x + у+7)2 +2 p(x? + ў +7) +4 p(p-1) y (x + y + 2)*² +2 p(x² + y + ?)*1 +4 p(p-1)z (x + y + 2)"?+2 p(x + ỷ + р-2 = 0 Which can simplified to [4 p(p-1) + 6 p](x² + y} + z)* = = 0. The above equation is true for x, y and z (except possibly at (0,0,0)) if the constant p is selected such that 4 p(p-1) + 6 p= 0= p(4 p+ 2) = 0= p=0, p=-1/2.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section: Chapter Questions
Problem 49RE
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hello, i am unable to understand this question even though i have the working solution, please advice me step by step. i dont even understand how did the first equation get formed, is there some formula ?

and = 4 p(p- 1)x² (x² + y+²)*² +2 p(x² + y° +2)P!, -= 4 p(p-1) y (x + ỷ +z')*² +2 p(x² + yỷ + ?)®, = 4 p(p-1) z² (x² + y² + z²)? +2 p(x° + y° + *). Thus, V²ø = 0 gives 4p(p-1)x (x + у+7)2 +2 p(x? + ў +7) +4 p(p-1) y (x + y + 2)*² +2 p(x² + y + ?)*1 +4 p(p-1)z (x + y + 2)"?+2 p(x + ỷ + р-2 = 0 Which can simplified to [4 p(p-1) + 6 p](x² + y} + z)* = = 0. The above equation is true for x, y and z (except possibly at (0,0,0)) if the constant p is selected such that 4 p(p-1) + 6 p= 0= p(4 p+ 2) = 0= p=0, p=-1/2.
Transcribed Image Text:and = 4 p(p- 1)x² (x² + y+²)*² +2 p(x² + y° +2)P!, -= 4 p(p-1) y (x + ỷ +z')*² +2 p(x² + yỷ + ?)®, = 4 p(p-1) z² (x² + y² + z²)? +2 p(x° + y° + *). Thus, V²ø = 0 gives 4p(p-1)x (x + у+7)2 +2 p(x? + ў +7) +4 p(p-1) y (x + y + 2)*² +2 p(x² + y + ?)*1 +4 p(p-1)z (x + y + 2)"?+2 p(x + ỷ + р-2 = 0 Which can simplified to [4 p(p-1) + 6 p](x² + y} + z)* = = 0. The above equation is true for x, y and z (except possibly at (0,0,0)) if the constant p is selected such that 4 p(p-1) + 6 p= 0= p(4 p+ 2) = 0= p=0, p=-1/2.
5. Find value(s) of the constant p such that ø(x, y, z) = (x + y + z)' satisfies Vø = 0 at all points (х,у,2) еxcept possibly at (0,0,0). Solution. V'ø = 0 can be written out in full as = 0. From ø (x, y, z) = (x² + y + z)®, we obtain = 2 px(x + y + z)P!, Əx = 2 py(x + y + z)*, ду 2 pz(x² + y} + z)P.
Transcribed Image Text:5. Find value(s) of the constant p such that ø(x, y, z) = (x + y + z)' satisfies Vø = 0 at all points (х,у,2) еxcept possibly at (0,0,0). Solution. V'ø = 0 can be written out in full as = 0. From ø (x, y, z) = (x² + y + z)®, we obtain = 2 px(x + y + z)P!, Əx = 2 py(x + y + z)*, ду 2 pz(x² + y} + z)P.
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