anonymous
  • anonymous
if both the roots of x^2-(p-4)x+2e^(2log p)-4=0 are negative then what are the possible range of values of p
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
x^2-(p-4)x+2P^2-4=0
anonymous
  • anonymous
if both are -ve so p-4<0 and p<4 also p>0 as it is in ln.
amistre64
  • amistre64
i was on the cusp of saying something similar :)

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anonymous
  • anonymous
but agin diccrim. will be +ve.
anonymous
  • anonymous
why should p-4 be less than zero if the roots are negative?
amistre64
  • amistre64
(r+m)(r+n) amkes the roots both -
anonymous
  • anonymous
(p-4)^2>4(2p^2-4) 7p^2+8p-32<0
amistre64
  • amistre64
+m+n = + result for -(p-4)
amistre64
  • amistre64
-p+4 >/ 0 -p >/ -4 p
amistre64
  • amistre64
2e^(2log p)-4 >/ 0 since +m(+n) = +result
anonymous
  • anonymous
not following you guys..:-(
anonymous
  • anonymous
what is your prob?????????
amistre64
  • amistre64
suppose we got: (r-4)(r-6) are the zeros here + or - ?
anonymous
  • anonymous
dipankar: u said "if both are -ve so p-4<0" why is this so?
anonymous
  • anonymous
because if ax^2+bx+c=0 has roots m,n then m+n=-b/a.
anonymous
  • anonymous
here b=-(p-4), a=1 and two roots are-ve so m+n=p-4 is -ve.
amistre64
  • amistre64
2e^(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
  • anonymous
ok...got it now.. so whats nxt?
amistre64
  • amistre64
ln(ln(0)/ln(10)) >/ 0 ln(ln(p)) - ln(10) >/ 0 i think this is how to do it :)
anonymous
  • anonymous
p<4 also p>0 is that the range?
anonymous
  • anonymous
no you also have to satisfy that D>=0
anonymous
  • anonymous
i.e. 7p^2+8p-32<0
anonymous
  • anonymous
if this gives some value of p that is less than 4 but>0 that will be your range.
amistre64
  • amistre64
ln(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
  • amistre64
ln(p)/ ln(10) >/ e^0 = 1 log(p) >/ 1 ....same result for me that way.....
anonymous
  • anonymous
what is your problem now?????????
anonymous
  • anonymous
a<0, b<0 a+b<0, ab>0 therefore p-4<0 => p<4; 2p^2-4>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^2-4ac>=0 (p-4)^2-4(2p^2-4)>=0 =>7p^2+8p-32<=0 therefore \[\sqrt{2}
anonymous
  • anonymous
yes..
anonymous
  • anonymous
what are the roots of the eqn 7p^2+8p-32<=0
anonymous
  • anonymous
can you tell me please.
anonymous
  • anonymous
(4(+-)sqrt(230))/7
anonymous
  • anonymous
how????????? its wrong.
anonymous
  • anonymous
(-8(+-)sqrt(960))/7
anonymous
  • anonymous
yea i factored 2 out
anonymous
  • anonymous
n yea it should be -ve 4 up there
anonymous
  • anonymous
hey how did u get ur denominator to be 7?? shouldnt it be 14??
anonymous
  • anonymous
sorry 14.
anonymous
  • anonymous
and also it sqrt(920) nt 960..
anonymous
  • anonymous
why? 32*4*7+64
anonymous
  • anonymous
yea..ryt..my bad
anonymous
  • anonymous
wrong!, what is all this noobery
anonymous
  • anonymous
it says both the roots are negative , not that they are complex
anonymous
  • anonymous
it says both the roots are negative , not that they are complex
anonymous
  • anonymous
if both the roots are negative that means that the sum must be negative, and the product must be positive

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