Question

See the attachment

Answer 1

@TuringTest

Answer 2

Will it work like this ?:
\[\large{x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{k+1}{k}}\]
\[\large{x_2 = \frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{k+2}{k+1}}\]

Answer 3

product of roots c/a= k+2/k
sum of roots -b/a=.......
simpler, but this may not work too.....

Answer 4

\(a^2(1-(-b/a)+(c/a))^2\)

Answer 5

but we don't need to find that ^ in terms of k.....

Answer 6

yep ..

Answer 7

it worked thanks..

Answer 8

"product of roots c/a= k+2/k
sum of roots -b/a=.......
simpler, but this may not work too....."

this worked for me :)

Answer 9

got k+1 as -a/sqrt(b^2-4ac) ??

Answer 10

This may help:
(k^2+k) x^2 -(2k^2+4k+1)x + (k^2 +3k+2)=0

Answer 11

So, (a+b+c)^2=(1)^2=1

Answer 12

Also, b^2 -4ac =1

Answer 13

So, (a+b+c)^2 =b^2-4ac

Answer 14

got it? @mathslover

Answer 15

correct? @hartnn

Answer 16

for b^2-4ac, u actually put a,b,c in terms of k and got 1 ?

Answer 17

Yep

Answer 18

a=k^2+k
b=-2k^2-4k-1
c=k^2+3k+2

Answer 19

then i want to know whether @mathslover had simpler solution

Answer 20

So, (a+b+c)^2 =b^2-4ac
is correct

Answer 21

yes,i think..... and your solution is also elegant but what did he do ?

Answer 22

bit busy right now..

will post solution later ...

Answer 23

OK

Answer 24

"product of roots c/a= k+2/k
sum of roots -b/a=.......
simpler, but this may not work too....."

this worked for me two :)

Answer 25

???

Answer 26

method started by hartnn in first reply... we can pursue.. .

from product of roots, c/a = (k+2)/k
=> k = 2a/(c-a) -------------(1)

from sum of roots, -b/a = 2 + 1/(k^2+k) ------(2)

substituting (1) in (2) gives,

(a+b+c)^2 =*****

but i didnt took it till the end... but i see its getting shaped well...

Answer 27

oh, yes ......maybe @mathslover did same thing....and this is much simpler....