Question

The sum of the roots of \(\LARGE{\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}}\) is zero. Find the product of the roots.

Answer 1

I did this through right here \[\frac{(x+b)+(x+a)}{x^2+(a+b)x+ab}=\frac{1}{c}\]\[{2x+a+b \over x^2+(a+b)x+ab}=\frac{1}{c}\]By cross multiplication\[2xc+ac+bc=x^2+(a+b)x+ab\]

Answer 2

I wanna solve this further

Answer 3

bring it in the form Ax^2+Bx+C=0
subtract left side equation.

Answer 4

looking good so far
now set it equal to zero

Answer 5

\[x^2+(a+b)x+c-2xc-ac-bc=0\]

Answer 6

x^2+(a+b-2c)+(ab-bc-ac)=0
sum of roots = -B/A =... ?

Answer 7

x^2+(a+b-2c)x+(ab-bc-ac)=0

Answer 8

-b/a=0

Answer 9

(a+b-2c)=...?

Answer 10

-b/a=0

Answer 11

But how did you put (a+b-2c) i mean from where do you put -2c

Answer 12

\(x^2+(a+b)x+ab-2xc-ac-bc=0 \\ x^2+(a+b-2c)x+(ab-bc-ac)=0 \\Ax^2-Bx+C=0\\-B/A=sum=0\implies a+b-2c=0\)

Answer 13

sorry, thats should be Ax^2+Bx+C=0

Answer 14

got it ?

Answer 15

yes

Answer 16

@hartnn what should we do further ??

Answer 17

so, \( a+b=2c\)
Product of roots \(= ab-c(a+b) = ab-c(2c) = ab-2c^2 \)

Answer 18

Thank god now it's fine :)

Answer 19

\[\Huge{\color{green}{\text{Ans.=}\frac{-1}{2}(a^2+b^2)}}\]

Answer 20

I'm nt getting this. few minutes ago i thought that i can gt it but i couldn't :(

Answer 21

\(ab-2c^2 = -1/2(4c^2-2ab )=-1/2((2c)^2)-2ab=-1/2((a+b)^2-2ab)\)
now ?

Answer 22

There's this pleasant thing called Vieta's Formula.

Answer 23

I'm unkown person from vieta's formula (Mein agyat vyakti hun, mujhe nahin pata vieta's formula kya hota hai ) :/

Answer 24

Ohh ! so u multiplied it by -2 isn't it ?

Answer 25

you can say multiplied and divided by -2
i've shown you all steps, if there's dout, ask.

Answer 26

No, this was the only doubt, thanx :)

Answer 27

welcome ^_^

Answer 28

\[x^2 + bx + c\]The sum of roots is \(-b\) and the product is \(c\). As a derivation,\[ax^2 + bx + c = a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right)\]The sum is \(-\dfrac{b}{a}\) and the product is \(\dfrac{c}{a}\).

Answer 29

What do u mean by derivation @ParthKohli ?

Answer 30

Dusra wala formula pehle waale formula se nikala gaya hai =)

Answer 31

Main bataun kaise nikalte hain ye formula ko?

Answer 32

haan

Answer 33

@ParthKohli

Answer 34

Achcha. Maan lo humare paas ek quadratic expression hai in the form \(x^2 + bx + c\). Suppose karo that the roots are \(s,r\). Toh \(x^2 + bx + c = (x - s)(x-r) = x^2 - (s +r)x + sr\). Coefficients equate karo toh \(b = -(s+r)\) aur \(sr = c\).

Answer 35

Toh yeh voh formula hai ??

Answer 36

ya is formulae ki tip hai ??

Answer 37

Haan, humne pehla formula derive kar liya.

Answer 38

Thike

Answer 39

Ab agar humare paas \(ax^2 + bx + c = 0\) hai toh hum dono sides se \(a\) divide kar sakte hain. Toh we have\[x^2 + \left(b \over a\right)x + \left(c \over a\right)\]Ispe pehle formula apply karo

Answer 40

apply kar toh liya

Answer 41

kya timepass chalu hai ?? :P

Answer 42

lol

Answer 43

Aap toh IIT se nikal liye hum bachchon ka bhi kuch karo yaar @hartnn :D

Answer 44

mera matlab hai tumne a divide toh kar liya hai now multiply it by a u will again gt x^2+bx+c=0

Answer 45

nikal liye??? abhi bhi hu !

Answer 46

Hain?

Answer 47

Bhaiya mujhe toh 5 sal hai iit ke liye. bhadiya tayyari ho jayegi :)

Answer 48

IIT enter karna mushkil hai! Main IIT nahi jaunga

Answer 49

Toh kahan jaoge

Answer 50

MIT ?

Answer 51

Ayashi karunga =)

Answer 52

LOL ;)

Answer 53

par mein toh yeh dusra formula derive nahi kar paa raha hun, iit kya khaak karunga

Answer 54

Arrey asaan hai

Answer 55

kaise ?

Answer 56

Dekho, humare paas yeh hai:\[x^2 + {b\over a}x + {c\over a}\]Right?

Answer 57

haan

Answer 58

Toh pehla wala formula kehta hai ki,\[\text{sum of roots} \implies -(\text{coefficient of }x) \\ \text{product of roots}\implies \text{coefficient of }x^0\]

Answer 59

haan itna toh gyat hai mujhe :)

Answer 60

Toh yeh equation mein sum of roots \(-\dfrac{b}{a}\) hai aur product hai \(\dfrac{c}{a}\) =)

Answer 61

haan

Answer 62

Aur ho gaya dusra formula derive!

Answer 63

bas itna hi hai :p

Answer 64

Yup!

Answer 65

Bada aasan tha, Dhanyavaad aap dono ko :)

Answer 66

Koi nahi, aap bhai ho apne =D

Answer 67

yeh to aapka badappan hai :)