## koli123able 2 years ago Find 'p' if 3x^2+2x+3p=0 has real & different roots. . . ANSWER = p<1/9

1. koli123able

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

2. Xavier

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

3. koli123able

delta?? u mean discriminant?

4. ParthKohli

Yes.

5. Xavier

Yup

6. koli123able

so how D>0 ??

7. Xavier

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

8. koli123able

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

9. harsimran_hs4

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

10. agent0smith

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 $b^2-4ac > 0$

11. koli123able

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

12. koli123able

am I right?

13. harsimran_hs4

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

14. harsimran_hs4

yes you are!!

15. koli123able

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

16. koli123able

is this ok?

17. Xavier

You lost a p at the end

18. koli123able

oh sorry I typed it in hurry

19. koli123able

p<1/9