Question

Please see attachment ..

Answer 1

Solve the equation

Answer 2

@mukushla @vishweshshrimali5

Answer 3

i think use dividendo & componendo

Answer 4

@mathslover :)

Answer 5

well right ... nice idea i am going to try that wait

Answer 6

k!

Answer 7

k! = 1*2*3*4*... *k :D

Answer 8

I meant OK

Answer 9

So i got the following .. equation (quadratic): also first root x = 0

7x^3-43x+40=0

we get : second root = 5
third root = 8 / 7

Answer 10

x2(2x-3)+(x+1)/x2(2x-3)-(x+1)=x2(3x-1)+(5x-13)/x2(3x-1)-(5x-13)

Answer 11

i m trying factorization @mathslover

Answer 12

hmn kkk

Answer 13

\[\qquad\quad\qquad\qquad\qquad\frac{2x^3-3x^2-x+1}{2x^3-3x^2-x-1}=\frac{3x^3-x^2+5x-13}{3x^3-x^2-5x+13}\]
\[\frac{2x^3-3x^2-x-1}{2x^3-3x^2-x-1}+\frac{2}{{2x^3-3x^2-x-1}}=\frac{3x^3-x^2+5x+13}{3x^3-x^2-5x+13}+\frac{-26}{3x^3-x^2-5x+13}\]
\[\qquad\qquad\qquad\qquad1+\frac{2}{{2x^3-3x^2-x-1}}=1+\frac{-26}{3x^3-x^2-5x+13}\]

Answer 14

I am right:

also solution 1 = 0

Answer 15

@jiteshmeghwal9 just try the quadratic equation formula there ... factorization will use the mind in a worst way .. :)

@UnkleRhaukus Why to do in such a long manner ? : as @jiteshmeghwal9 told .. use componendo and dividendo ..

Answer 16

so the method of Dividendo & componendo was also right

Answer 17

so my N.T.S.E. works @ last haha :)

Answer 18

and further we will get the quadratic equation by having the first soln as 0

@UnkleRhaukus

Answer 19

yes @jiteshmeghwal9 good work .. thanks

thanks @UnkleRhaukus also

Answer 20

dividendo & componendo ?

Answer 21

yw:)

Answer 22

crazy looking words

Answer 23

yes @UnkleRhaukus

if a/b = c/d then :

(a+b)/(a-b)=(c+d)/(c-d) only if a is not equal to b and c is not equal to d

Answer 24

huh

Answer 25

that is in LaTeX :
\[\large{\textbf{IF} a\ne b \space \textbf{and}\frac{a}{b}=\frac{c}{d}}\]
\[\large{\textbf{THEN}\frac{a+b}{a-b}=\frac{c+d}{c-d}}\]

Answer 26

@UnkleRhaukus you want the proof of that?

Answer 27

news to me,
yeah ok,

Answer 28

http://www.meritnation.com/discuss/question/683600

Answer 29

I was tired writing the latex sorry @UnkleRhaukus

Answer 30

dividendo & componendo, interesting

Answer 31

this is the proof ..

(inverse)

Answer 32

@UnkleRhaukus That may help u in understanding componendo and dividendo