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Discriminant Related Question

  • Thread starter _Mayday_
  • Start date
  • #1
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[SOLVED] Discriminant Related Question

Question

[tex]x^2 - 8x - 29 = (x+a)^2 +b[/tex], where a and b are constant.

NOTE: The equals sign should be an always equals sign, so like three lines under each other :bugeye:

(a) Find the value of a and b

(b)Hence, or otherwise show the roots of [tex]x^2 - 8x - 29 = 0[/tex] are [tex]c =+/- d\sqrt5[/tex] where c and d are integers.

Attempt

(a)[tex]x^2 - 8x - 29 = (x+a)^2 +b[/tex]

[tex](x-4)^2 - 16 - 29 = (x+a)^2 +b[/tex]

a = -4
b = -45

(b) Erm...not sure, as it says roots I will assume it means two roots so [tex]b^2 - 4ac > 0[/tex]

64 + 116 > 0, yes this is all alright but I am not sure how to get it in the form [tex]c =+/- d\sqrt5[/tex]

_Mayday_
 

Answers and Replies

  • #2
NOTE: The equals sign should be an always equals sign, so like three lines under each other :bugeye:
Oh that's done with \equiv

[tex]\equiv[/tex]
 
  • #3
798
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Oh that's done with \equiv

[tex]\equiv[/tex]
Cheers!

Anyone got any ideas on how to solve the other question? :rofl:
 
  • #4
HallsofIvy
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What are the roots of the equation x2- 8x- 29= 0? Since you have already "completed the square" that should be easy!
 
  • #5
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By find the roots do they just mean find the possible values for x? The question is what has got me, what do they mean by find the roots? I know I should know this but, I think I have already done the 'hard' bit.
 
  • #6
798
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[tex](x-4)^2 = 45[/tex]

[tex]\sqrt{45} = +/- 3\sqrt5[/tex]
 
  • #7
798
0
[tex]x - 4 = +\- 3\sqrt5[/tex]

[tex]x = 4 +/- 3\sqrt5[/tex]
 
  • #8
798
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I think that is correct, thank you!
 

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