Question

If p(q-r)x^2 +q(r-p)x + r(p-q) has equal roots,then, 2/q=??
a)p+ 1/r c)p+r
b) 1/p +r d) 1/p + 1/r

Answer 1

@nitz

Answer 2

\[D = b^2 - 4ac\]

For equal roots D = 0

Answer 3

i know that, tell me what is 2/q and how ??

Answer 4

cosidering symmtry of that qn ans wud be d :P

Answer 5

It is the value you have to find from this approach...

Answer 6

just put those things in nd solve....u l get it

Answer 7

got your answer?

Answer 8

no

Answer 9

i have solved it till
pqr(p-q)-pr^2(p-q)=0

Answer 10

wait ill check

Answer 11

i am sorry i am not getting answer now
will tell you in the evening

Answer 12

ok

Answer 13

any other problem in quadratic?

Answer 14

yes.

Answer 15

wait.

Answer 16

ok

Answer 17

Can you show me how have you arrived at that equation??

Answer 18

just simplified everything.

Answer 19

I want to see if you want to show me..

Answer 20

\[q^{2}r^{2}-2pq^{2}r+p^{2}r^{2}-4pq(pr-qr)-pr(pr-qr)=0\]

Answer 21

this came by putting D=0.

Answer 22

How you have put the value of 4ac??

Answer 23

\[b^2 - 4ac = 0\]

\[q^2(r-p)^2 - 4pr(p-q)(q-r) = 0\]

Answer 24

yes.

Answer 25

\[ q^2(r-p)^2-4pr(q-r)(p-q)=0\]
make this a quadratic of \( q^2 \)
First, expand the 2nd term to get
\[ -4pr(pq-pr-q^2+qr)= -4p^2rq+4p^2r^2+4pq^2r-4pqr^2 \]
\[= 4pr q^2 -4pr(p+r) q +4p^2r^2 \]
this will make the expression

\[ ((r-p)^2+4pr) q^2 -4pr(p+r) q + 4p^2r^2=0 \]
the coefficient of q^2 can be simplified to (p+r)^2
\[ (p+r)^2 q^2 -4pr(p+r) q + 4p^2r^2=0 \]

solve for q using the quadratic formula
\[ \frac{4pr(p+r)}{2(p+r)^2} \pm \frac{\sqrt{D}}{2(p+r)^2} \]
where D is
\[ 16p^2r^2(p+r)^2 - 16(p+r)^2 p^2r^2=0 \]
we find q
\[ q= \frac{2pr}{(p+r)} \]
and
\[ \frac{2}{q}= \frac{p+r}{pr}= \frac{1}{r}+\frac{1}{p} \]