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

a) Find the value of p.

b) for this value of p, solve the equation.

2. Greetings,

First of all, think when quadratic equations have equal roots.
From there you get two values for p, one of which is negative and solving the equation with the other p is not a problem, I shall assume.

3. Originally Posted by randomised

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

a) Find the value of p.

b) for this value of p, solve the equation.
An equation has repeating [equal] roots when the discriminant is equal to zero.

$\Delta=b^2-4ac\implies \Delta=(2p)^2-4(1)(3p+4)=4p^2-12p-16$

Thus, $\Delta=0$ when $4p^2-12p-16=0\implies p^2-3p-4=0$

Can you take it from here and solve for $p$?

--Chris

4. Originally Posted by randomised

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

a) Find the value of p.

b) for this value of p, solve the equation.
Set the discriminant to zero for a double root.

$b^2-4ac=0$

a = 1

b = 2p

c = 3p + 4
Geeze! Way too slow today!

5. Perhaps D = k^2 - ac, k = b/2 would shorten the path.