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anonymous
 5 years ago
if both the roots of x^2(p4)x+2e^(2log p)4=0 are negative then what are the possible range of values of p
anonymous
 5 years ago
if both the roots of x^2(p4)x+2e^(2log p)4=0 are negative then what are the possible range of values of p

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if both are ve so p4<0 and p<4 also p>0 as it is in ln.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i was on the cusp of saying something similar :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but agin diccrim. will be +ve.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why should p4 be less than zero if the roots are negative?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(r+m)(r+n) amkes the roots both 

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(p4)^2>4(2p^24) 7p^2+8p32<0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0+m+n = + result for (p4)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0p+4 >/ 0 p >/ 4 p </ 4

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02e^(2log p)4 >/ 0 since +m(+n) = +result

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not following you guys..:(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is your prob?????????

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0suppose we got: (r4)(r6) are the zeros here + or  ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dipankar: u said "if both are ve so p4<0" why is this so?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because if ax^2+bx+c=0 has roots m,n then m+n=b/a.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here b=(p4), a=1 and two roots areve so m+n=p4 is ve.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02e^(2log p)4 >/ 0 2e^(2log p) >/ 4 e^(2log p) >/ 2 ln(2 log(p)) >/ ln(2) ln(2) + ln(log(p)) >/ ln(2) ln(log(p)) >/ 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok...got it now.. so whats nxt?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ln(ln(0)/ln(10)) >/ 0 ln(ln(p))  ln(10) >/ 0 i think this is how to do it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0p<4 also p>0 is that the range?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no you also have to satisfy that D>=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if this gives some value of p that is less than 4 but>0 that will be your range.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ln(ln(0)/ln(10)) >/ 0 ln(p)/ ln(10) >/ e^0 = 1 ln(p) >/ ln(10) p >/ 10 ok, what if anything did i do wrong

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ln(p)/ ln(10) >/ e^0 = 1 log(p) >/ 1 ....same result for me that way.....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is your problem now?????????

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a<0, b<0 a+b<0, ab>0 therefore p4<0 => p<4; 2p^24>0 =>p^2>2 therefore p>sqrt{2} or, p<sqrt(2) so, p\[p \in(\sqrt{2},4) [p>0]\] again both roots are ve. b^24ac>=0 (p4)^24(2p^24)>=0 =>7p^2+8p32<=0 therefore \[\sqrt{2}<p<(4/7)(\sqrt{15}1)\] did i get it right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what are the roots of the eqn 7p^2+8p32<=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you tell me please.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how????????? its wrong.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0n yea it should be ve 4 up there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey how did u get ur denominator to be 7?? shouldnt it be 14??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and also it sqrt(920) nt 960..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wrong!, what is all this noobery

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it says both the roots are negative , not that they are complex

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it says both the roots are negative , not that they are complex

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if both the roots are negative that means that the sum must be negative, and the product must be positive
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