Question

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

Answer 1

x^2-(p-4)x+2P^2-4=0

Answer 2

if both are -ve so p-4<0
and p<4 also p>0 as it is in ln.

Answer 3

i was on the cusp of saying something similar :)

Answer 4

but agin diccrim. will be +ve.

Answer 5

why should p-4 be less than zero if the roots are negative?

Answer 6

(r+m)(r+n) amkes the roots both -

Answer 7

(p-4)^2>4(2p^2-4)
7p^2+8p-32<0

Answer 8

+m+n = + result for -(p-4)

Answer 9

-p+4 >/ 0

-p >/ -4

p </ 4

Answer 10

2e^(2log p)-4 >/ 0 since +m(+n) = +result

Answer 11

not following you guys..:-(

Answer 12

what is your prob?????????

Answer 13

suppose we got:

(r-4)(r-6) are the zeros here + or - ?

Answer 14

dipankar: u said "if both are -ve so p-4<0"
why is this so?

Answer 15

because if ax^2+bx+c=0 has roots m,n then m+n=-b/a.

Answer 16

here b=-(p-4), a=1 and two roots are-ve so m+n=p-4 is -ve.

Answer 17

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

Answer 18

ok...got it now.. so whats nxt?

Answer 19

ln(ln(0)/ln(10)) >/ 0

ln(ln(p)) - ln(10) >/ 0 i think this is how to do it :)

Answer 20

p<4 also p>0 is that the range?

Answer 21

no you also have to satisfy that D>=0

Answer 22

i.e. 7p^2+8p-32<0

Answer 23

if this gives some value of p that is less than 4 but>0 that will be your range.

Answer 24

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

Answer 25

ln(p)/ ln(10) >/ e^0 = 1

log(p) >/ 1 ....same result for me that way.....

Answer 26

what is your problem now?????????

Answer 27

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 \[\sqrt{2}<p<(4/7)(\sqrt{15}-1)\]

did i get it right?

Answer 28

yes..

Answer 29

what are the roots of the eqn 7p^2+8p-32<=0

Answer 30

can you tell me please.

Answer 31

(4(+-)sqrt(230))/7

Answer 32

how????????? its wrong.

Answer 33

(-8(+-)sqrt(960))/7

Answer 34

yea i factored 2 out

Answer 35

n yea it should be -ve 4 up there

Answer 36

hey how did u get ur denominator to be 7?? shouldnt it be 14??

Answer 37

sorry 14.

Answer 38

and also it sqrt(920) nt 960..

Answer 39

why? 32*4*7+64

Answer 40

yea..ryt..my bad

Answer 41

wrong!, what is all this noobery

Answer 42

it says both the roots are negative , not that they are complex

Answer 43

it says both the roots are negative , not that they are complex

Answer 44

if both the roots are negative that means that the sum must be negative, and the product must be positive