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The equation x2+2px+(3p+4)=0, x^2 + 2px + (3p + 4) = 0, where P P is a positive constant, has equal roots.

a) Find the value of P P

b=2P,a=1,c=(3P+4)[br]Equalroots=b24ac=0[br](2p)24x(1)(3P+4)[br]4p212p16=0./4[br]P23p4=0[br](P+1)(P4)=0[br]P=1[i]orP=4[/i] b = 2P, a = 1, c = (3P + 4)[br]Equal roots = b^2 - 4ac = 0[br](2p)^2 - 4 x (1)(3P + 4)[br]4p^2 - 12p - 16 = 0. /4 [br]P^2 - 3p - 4 = 0 [br](P + 1) (P - 4)=0 [br]P = -1 [i]or P = 4 [/i]

b) For this value of P P , solve the equation x2+2px+(3p+4)x^2 + 2px + (3p + 4)

How do I decide which value of P to use :s-smilie: sorry about the weird question marks above, don't know how to remove them!
Look at the information given about P in the question :smile:

Original post by animemaniac132

How do I decide which value of P to use :s-smilie: sorry about the weird question marks above, don't know how to remove them!
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Thanks

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