Question

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

Answer 1

I have no idea on solving this question....

Answer 2

A quadratic has real and different roots if the descriminant is greater than zero: \[\Delta > 0\]

Answer 3

delta?? u mean discriminant?

Answer 4

Yes.

Answer 5

Yup

Answer 6

so how D>0 ??

Answer 7

What do you mean? Set the discriminant to be greater than zero
b^2-4ac > 0

Answer 8

I mean how is discriminant greater than zero can u explain?

Answer 9

if d > 0 then there are real and different roots
d = 0 real and equal roots
d < 0 imaginary roots

Answer 10

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 \[b^2-4ac > 0 \]

Answer 11

If D>0 then sqrt(b^2-4ac) is a real number and roots are real and unequal

Answer 12

am I right?

Answer 13

\[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\]
try putting different values of d and check out

Answer 14

yes you are!!

Answer 15

as D>0 and D=b^2-4ac
so, (2)^2-4*3*3p>0
4-36p>0
4>36p
1/9>0

Answer 16

is this ok?

Answer 17

You lost a p at the end

Answer 18

oh sorry I typed it in hurry

Answer 19

p<1/9