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koli123able
 2 years ago
Best ResponseYou've already chosen the best response.0I have no idea on solving this question....

Xavier
 2 years ago
Best ResponseYou've already chosen the best response.2A quadratic has real and different roots if the descriminant is greater than zero: \[\Delta > 0\]

koli123able
 2 years ago
Best ResponseYou've already chosen the best response.0delta?? u mean discriminant?

Xavier
 2 years ago
Best ResponseYou've already chosen the best response.2What do you mean? Set the discriminant to be greater than zero b^24ac > 0

koli123able
 2 years ago
Best ResponseYou've already chosen the best response.0I mean how is discriminant greater than zero can u explain?

harsimran_hs4
 2 years ago
Best ResponseYou've already chosen the best response.0if d > 0 then there are real and different roots d = 0 real and equal roots d < 0 imaginary roots

agent0smith
 2 years ago
Best ResponseYou've already chosen the best response.0Real 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 \[b^24ac > 0 \]

koli123able
 2 years ago
Best ResponseYou've already chosen the best response.0If D>0 then sqrt(b^24ac) is a real number and roots are real and unequal

harsimran_hs4
 2 years ago
Best ResponseYou've already chosen the best response.0\[x = \frac{ b \pm \sqrt{b^2  4ac} }{ 2a }\] try putting different values of d and check out

koli123able
 2 years ago
Best ResponseYou've already chosen the best response.0as D>0 and D=b^24ac so, (2)^24*3*3p>0 436p>0 4>36p 1/9>0

koli123able
 2 years ago
Best ResponseYou've already chosen the best response.0oh sorry I typed it in hurry
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