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Find 'p' if 3x^2+2x+3p=0 has real & different roots. . . ANSWER = p<1/9

Mathematics
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I have no idea on solving this question....
A quadratic has real and different roots if the descriminant is greater than zero: \[\Delta > 0\]
delta?? u mean discriminant?

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Other answers:

Yes.
Yup
so how D>0 ??
What do you mean? Set the discriminant to be greater than zero b^2-4ac > 0
I mean how is discriminant greater than zero can u explain?
if d > 0 then there are real and different roots d = 0 real and equal roots d < 0 imaginary roots
Real and different...? I'm guessing you mean real and distinct. But yes you can use the discriminant. When the discriminant is less than zero, it will have complex/imaginary roots, as the quadratic formula takes the square root of the discriminant ( http://www.purplemath.com/modules/quadform.htm ) \[3x^2+2x+3p=0 \]\[ax^2+bx+c = 0\] so here, a=3, b = 2 and c = 3p discriminant is \[b^2-4ac > 0 \]
If D>0 then sqrt(b^2-4ac) is a real number and roots are real and unequal
am I right?
\[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\] try putting different values of d and check out
yes you are!!
as D>0 and D=b^2-4ac so, (2)^2-4*3*3p>0 4-36p>0 4>36p 1/9>0
is this ok?
You lost a p at the end
oh sorry I typed it in hurry
p<1/9

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