Question

Question based on nature of roots

Answer 1

If one root of (l-m)x^2 +lx +1 =0 be double the other and if l be real , show that
\[m \le \frac{ 9 }{ 8 }\]
Now how the hell i am supposed to show that .

Answer 2

I is a variable, right?

Answer 3

Natural log

Answer 4

\(x_1=2x_2\)
\(f(x_1)=f(2x_2)=0\)

Answer 5

yes

Answer 6

\(\large (l-m){x_1}^2+lx_1+1= (l-m){x_2}^2+lx_2+1\)

Answer 7

\(\large (l-m){2x_2}^2+l2x_2+1= (l-m){x_2}^2+lx_2+1\)

Answer 8

I disagree.
\[(l-m)(2a)^2+l(2a)+1=(l-m)a^2+la+1\]

Answer 9

Assuming that a and 2a are the 2 roots

Answer 10

disagree what lol ?

Answer 11

(2a)^2 is not the same as 2a^2

Answer 12

@science0229 ive got what u mean its only a typo
check the steps and ull know :)

Answer 13

I m confused a bit

Answer 14

\(\large (l-m)\color{\red}{({2x_2)}^2}+l2x_2+1= (l-m){x_2}^2+lx_2+1\)

Answer 15

yes

Answer 16

\(\large (l-m){4x_2}^2+l2x_2+1= (l-m){x_2}^2+lx_2+1 \)

Answer 17

\(\large (l-m){3x_2}^2+lx_2=0\)

Answer 18

\(\large (l-m){3x_2}^2+lx_2=0 \)
\(\large x_2 [(l-m)3x_2+l]=0 \)

Answer 19

Wait are we proceeding to the answer

Answer 20

well lets find out :P

Answer 21

3(l-m)x_2+l=0

Answer 22

Is that a typo or what

Answer 23

l is a Natural log
then l > 0

Answer 24

Yes

Answer 25

no not a typo
\(\large x_2 [(l-m)3x_2+l]=0 \)
x_2 =0
or
\(\large [(l-m)3x_2+l]=0\)

Answer 26

yes k

Answer 27

solve for m

Answer 28

should i expand or what i should do

Answer 29

\(\large l-m=\frac{-l}{3x_2}\)

Answer 30

l - (-1)/3x2 =m

Answer 31

how does that QED

Answer 32

that doesnt :)

Answer 33

so r we incorrect?

Answer 34

no lol we need to continue :)

Answer 35

ok since its given there is real root then
b^2-4ac>0
l^2-2(l-m)>0
l^2>2(l-m)
-l^2/2 +l <m

Answer 36

yes

Answer 37

Yes m is greater than 0

Answer 38

so we have