1. Jan 8, 2008

### thomas49th

The equation $$x^{2} + 2px + (3p + 4) = 0$$, where p is a postitve constant, has equal roots

Q) Find the value of p

Well if it has equal roots b² - 4ac = 0
so

(2p)² - 4(1)(3p + 4) = 0

4p² - 12p - 16 = 0

4(p² - 3p - 4) = 0

4((p+1)(p-4)) = 0

so the anwser is p = 16

BUT the question booklet says p = 4

Which is right???
Thx :)

2. Jan 8, 2008

### d_leet

How did you get 16? Also there are two values of p that make the statement true.

3. Jan 8, 2008

### thomas49th

yeh i know the there are 2 valeus but p is a positive constant, as it says in the question

erm i got p = 4
but then u get the 4 otside the brackets so i multiply p by 4 giving me p = 16 or do you not count the four outside the brackets?

4. Jan 8, 2008

### Vid

Well, you can either divide both sides by 4 first to get p = 4 or if you wanted to keep the 4.
4(p-4) =0
4p-16 = 0
p = 4

5. Jan 8, 2008

### thomas49th

that was a stupid mistake on my part
sorry
yeh it was soooo simple

thanks
:)