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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

(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_

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# Homework Help: Discriminant Related Question

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