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Can any help with this As level maths core 1 quadratic function question?

I'm a little bit stuck on this question and was wondering if anyone can help.
The question is:
'The equation x^2 + 2px + (3p+4) =0 where p is a positive constant, has equal roots. Find the value of p (4 marks)'
So far I have this:
a=1 b=2p c=(3p+4). b^2 -4ac. (The discriminant)
(2p)^2 - (4 x 1 x (3p+4)
4p^2 = (4 x 1 x (3p+4)
4p^2 =12p +16
p^2= 3p + 4

But now I am not sure what to do, any help will be greatly appreciated
Thanks in advance
All you have to do now is solve the quadratic by factorisation:

p2−3p−4=0[br](p−4)(p+1)=0[br]p=4aspisapositiveconstant. p^2-3p-4=0[br](p-4)(p+1)=0[br]p=4 as p is a positive constant.
p^2 - 3p - 4 = 0
(p - 4)(p + 1) = 0

Since it's positive p = 4

Ninja'd ;o;
(edited 7 years ago)
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indigo2011
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Original post by NeverLucky
All you have to do now is solve the quadratic by factorisation:

p2−3p−4=0[br](p−4)(p+1)=0[br]p=4aspisapositiveconstant. p^2-3p-4=0[br](p-4)(p+1)=0[br]p=4 as p is a positive constant.


Thank you... Turns out I was factorising it wrong when I tried
Reply 4
Avatar for indigo2011
indigo2011
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Original post by Imperion
p^2 - 3p - 4 = 0
(p - 4)(p + 1) = 0

Since it's positive p = 4

Ninja'd ;o;


Thank you

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