Question

If the roots of the equation x^2 + px + q = 0 differ from the roots of the equation x^2+qx+p=0 by the same quantity, then (p is not equal to q) -
a) p+q+1=0 c) p+q+4=0
b)p+q+2=0 d) None of these

Answer 1

whats the ques ?? o.O

Answer 2

@shubhamsrg see on top

Answer 3

it says "then .."
then what ?

Answer 4

then you have 4 options

Answer 5

hmm,,what are the options supposed to be equal to ?
then what?
find the value of what? eh? you getting me?

Answer 6

sorry
all options = 0

Answer 7

using @apoorvk 's work,
(p+q)(p-q) = -4(p-q)
since p is not equal to q, we can cancel out (p-q)
=> p+q = -4
=> b)

Answer 8

oh so sorry there were a few mistakes up there in the first one.
hmm..



Let the first quadratic's roots be 'a' and 'b', and second quadratic's roots be 'c' and 'd'

So, \(|a-b|=|c-d|\)

For the sake of simplicity let me consider \(a\) and \(c\) as the greater of the roots respectively.
So,
\(a−b=c−d\)

Now, \(a−b=\sqrt{(a−b)^2}=\sqrt{(a+b)^2−4ab}=\sqrt{(−p)^2−4(q)}=\sqrt{p^2−4q}\)

Similarly, \(c−d=\sqrt{q^2−4p}\)

Now, a-b and c-d are equal.
So,

\(\sqrt{q^2−4p} = \sqrt{p^2−4q}\)


(though it wouldn't have made a difference really)
Now...
.
.
.

Answer 9

@shubhamsrg how (p+q)(p-q) = -4..................??

Answer 10

since p^2 - q^2 = 4(q-p)
=>(p+q)(p-q) = -4(p-q)

Answer 11

how p^2-q^2 = 4(q-p)??
tell me after- (p+2q)(p-2q) = (q+2p)(q-2p)

Answer 12

@nitz help me with the question i had sent to you.

Answer 13

ya i was solving it........

Answer 14

ok

Answer 15

@nitz you can that here.

Answer 16

*solve

Answer 17

shubham word me solve kiya hai .attach kar rahin hun see it

Answer 18

ok

Answer 19

@nitz but x+1>0 or x>-1
also the ans. has (-1,0)U(1,infinity)

Answer 20

now this was domain of function

Answer 21

if we see the inequality
(x+1/x-1)>2^0
(x+1/x-1)>1
x+1>x-1

Answer 22

Have you removed the log here??

Answer 23

yes

Answer 24

but how??
the log is only in numerator not in denominator.

Answer 25

\[\log_{2}(x+1)\div x-1 > 0\]

Answer 26

when did you tell that

Answer 27

again i have to solveeee

Answer 28

i've written it in the message sent to you.

Answer 29

Sorry

Answer 30

no it seemed to be over entire part

Answer 31

ok

Answer 32

ya got it

Answer 33

@nitz what happened?

Answer 34

sorry light chalee gayee thee

Answer 35

koi baat nhi.
Its India!!

Answer 36

case 1:
when x-1>0
ie when x>1
then,log(base 2)(x+1)>0
x+1>2^0
x+1>1
x>0
that is when x>1 then x>0..................(A)

Answer 37

ok

Answer 38

case 2:
when x-1<0 ie x<1
log(base 2)(x+1)<0....................sign is changed
x+1<1
x<0
now when x<1 then x<0

Answer 39

also here domain is x+1>0 ie x>-1

Answer 40

log(base 2)(x+1)<0....................sign is changed ??

Answer 41

when denominator is negative,sign gets reversed

Answer 42

ok

Answer 43

so combining all these conditions,you get your answer

Answer 44

the commom interval is (-1,0)U(1,inf)

Answer 45

ok thanks

Answer 46

answer aa gaya

Answer 47

trying!!

Answer 48

ok tell me where you dont get??

Answer 49

ok
my combined answer is not coming.

Answer 50

@nitz i am getting x =(-1,0)U(0,1)U(1,inf.)

Answer 51

see in first case the common interval will be \[(1,\infty) \]

Answer 52

right?

Answer 53

yes

Answer 54

ok

Answer 55

ok got it.
I had taken the intersection of all 5 conditions.

Answer 56

in second case the common interval was to be \[(-\infty,0) \]
but since x>-1
so common interval is (-1,0)

Answer 57

so (-1,0)U(1,inf)

Answer 58

interested in quadratics??

Answer 59

i got some problems there too.