solve {1/(a+b+x)}=(1/a)+(1/b)+(1/x), (a+b)not= 0

prakash kumar   11 years AGO

1/(a+b+x)=1/a + 1/b + 1/x
1/(a+b+x)=(bx+ax+ab)/abx
abx=abx+a2x+a2b+b2x+abx+ab2+bx2+ax2+abx
ax2+bx2+a2x+abx+abx+b2x+a2b+ab2=0
x2(a+b)+ax(a+b)+bx(a+b)+ab(a+b)=0
(a+b)(x2+ax+bx+ab)=0
since, a+b!=0
so, x2+ax+bx+ab=0
x(x+a)+b(x+a)=0
(x+a)(x+b)=0
x=-a,-b

Like? Yes (211) | No (68)  

Kumar Ankit   11 years AGO

( 1/ [a+b+x] ) = (1/a) +( 1/b )+ (1/x)

( 1 / [a+b+x] ) - (1 / x ) = (1/a) + (1/b)

=> { x - [a+b+x] } / ([a+b+x] * x ) = {a+b} / ab

=> - { a+b} / ( [a+b+x] * x ) = { a+b} / ab

=> -1 / ( [a+b+x] * x ) = 1 / ab

Cross Multiply : -ab = [a+b+x] * x

On Simplification : x2 + (a+b) x + ab = 0

Applying the Splitting the middle term method :

=> x2 + (a+b) x + ab = 0

=> [x2 + ax]+ [b x + ab] = 0

So : x ( x + a) + b ( x + a) = 0

=> (x + a ) * ( x + b ) = 0

Therefore : (x+a) = 0 or (x+b) = 0

Now : x = -a or x = -b
The value of x is : -a or -b

Like? Yes (110) | No (30)  

Grishma Shah   11 years AGO

1/a+b+x=1/a+1/b+1/x
=>(1/a+b+x)-1/x=1/a+1/b
Taking (a+b+x)x as LCM,
=> x-(a+b+x)/(a+b+x)x=b+a/ab
=> -(a+b)/x^2+ax+bx=a+b
Cancelling (a+b) from both sides we get,
=> -1/x2+ax+bx=1/ab
=> x^2+ax+bx= -ab
=> x^2+ax+bx+ab=0
=> x(x+a)+b(x+a)=0
=> x+b=0 or x+a=0
=> x=-b or x=-a

Like? Yes (93) | No (27)  

kumar sundram   11 years AGO

{1/(a+b+x)}=(1/a)+(1/b)+(1/c)
or, {1/(a+b+x)}={(bx+ax+ab)/abx}
or, abx=(a+b+x)(ax+bx+ab)
or, xa^2+abx+ba^2+abx+xb^2+ab^2+ax^2+bx^2+abx-abx=0
or, x(a^2+b^2+2ab)+ab(a+b)+x^2(a+b)=0
or, x(a+b)^2+ab(a+b)+x^2(a+b)=0
or, (a+b){x(a+b)+ab+x^2}=0
or, (a+b)(xa+xb+ab+x^2)=0
or, (a+b){x(a+x)+b(a+x)}=0
or, (a+b)(a+x)(x+b)=0
since, (a+b)not=0
so, (x+a)=0; i,e. x=-a
or, (x+b)=0; i,e. x=-b.

Like? Yes (63) | No (32)  

prabhat kumar   11 years AGO

1/(a+b+x)=1/a + 1/b + 1/x

1/(a+b+x)=(bx+ax+ab)/abx

abx=(a+b+x)(bx+ax+ab)

abx=(a+b+x){(a+b)x+ab)

abx=(a+b)x*a+(a+b)x*b+(a+b)*x^2 + a^2*b+a*b^2+abx

ax(a+b)+bx(a+b)+(a+b)x^2+ab(a+b)=0

x^2+ax+bx+ab=0

x(x+a)+b(x+a)=0

(x+a)(x+b)=0

hence,x=-a,x=-b

Like? Yes (40) | No (20)  

Tanvir Rahman   11 years AGO

1/(a+b+x)=i/a + 1/b + 1/x

1/(a+b+x)=(bx+ax+ab)/abx

abx=(a+b+x)(bx+ax+ab)

abx=(a+b+x){(a+b)x+ab)

abx=(a+b)x*a+(a+b)x*b+(a+b)*x^2 + a^2*b+a*b^2+abx

ax(a+b)+bx(a+b)+(a+b)x^2+ab(a+b)=0

x^2+ax+bx+ab=0

x(x+a)+b(x+a)=0

(x+a)(x+b)=0

hence,x=-a,x=-b

Like? Yes (31) | No (17)  

bollimuntha ravi teja   11 years AGO

1/ a+b+x = 1/a +1/b +1/x
1/a+b+x - 1/x = 1/a +1/b
L.C.M on both the sides
x-a-b-x/x(a+b+x) = a +b/ab
-a-b/ax +bx +x2 = a + b/ab
-(a+b)/ax+ bx+ x2 =a +b/ab
-1/ax + bx +x2 = 1/ab
cross multiply
-ab = ax + bx + x2
0 = x2 + ax + bx + ab
0 = x(x +a) +b(x + a)
0 = (x +a)(x + b)
x = -a , x = -b

Like? Yes (18) | No (12)  

VIVEK SINHA   11 years AGO

after solving quadratic egn. :
x1=(-(a+b)^2+a^2-b^2)/2(a+b)
x1=-b
similarly
x2=(-(a+b)^2-a^2+b^2)/2(a+b)
x2=-a
therefor x=-a,x=-b and correct optiion is 4

Like? Yes (18) | No (13)  

Dharmendra Kumar Rajak   11 years AGO

given eqn is {1/(a+b+x)}=(1/a)+(1/b)+(1/x),
and(a+b)not=0
after solving the eqn will be as-
=> x(a+b+x)+ab=0
=> x^2+ax+bx+ab=0
=> x(x+a)+b(x+b)=0
=> (x+a)(x+b)=0
=> x=-a,and x=-b;

Like? Yes (20) | No (18)  

Aman Kohli   11 years AGO

solving using options we will get that only fourth option will satisfy the equation.......

for x=-a

l.h.s= 1/(a+b+(-a)) = 1/b
r.h.s= (1/a) + (1/b) + (-1/a) = 1/b

l.h.s = r.h.s

for x=-b

l.h.s= 1/(a+b+(-b)) = 1/a
r.h.s= (1/a) + (1/b) + (-1/b) = 1/a

l.h.s = r.h.s

thus this is the ans (4).

Like? Yes (8) | No (6)  

amar nath tiwari   11 years AGO

answer@ 4
1/(x+a+b)=(ab+bx+ax)/abx
aftr solv it u got
(a+b)((x^2+x(a+b)+ab))=o
(a+b)=o
or
(x^2+x(a+b)+ab)=o
but given
a+b!= 0
so
(x^2+x(a+b)+ab)=0
by shree dharacharya rule
x=-a
x=-b
so answer
4

Like? Yes (10) | No (9)  

ABHISHEK CHANDEL   11 years AGO

{1/(a+b+x)}=(1/a)+(1/b)+(1/x),
abx=x(a+b)^2+x^2(a+b)+ab(a+b+x)
-ab=x(a+b)+x^2
x^2+x(a+b)+ab=0
now this is a quadratic eqn solve this eqn ans get x=-b&-a

Like? Yes (14) | No (13)  

Harjit kaur   11 years AGO

{1/(a+b+x)}=(1/a)+(1/b)+(1/x), (a+b)not= 0
taking LCM and equalizing we get
abx=(a+b+x)(bx+ax+ab)
={(a+b)+x}{(b+a)x+ab}
=(a+b)^2+(a+b)ab+(a+b)x^2+abx
(a+b)x^2+(a+b)^2)x+ab(a+b)=0
since (a+b)not = 0
So x^2+ax+bx+ab=0
x(x+a)+b(x+a)=0
(x+b)(x+a)=0
so x=-a,x=-b

Like? Yes (12) | No (12)  

Dinesh thatti   11 years AGO

if we LCM above eqn

x-a-b-x/x(a+b+x)=a+b/ab

solving this we get


-1/ax+bx+x^2=1/ab

cross multiply

-ab=ax+bx+x^2


x^2+ax+bx+ab=0

x(x+a)+b(x+a)=0

(x+a)(x+b)=0

x=-a , x= -b

Like? Yes (8) | No (9)  

SWETA YADAV   11 years AGO

ON solving the above equation we have a quadratic equation
(a+b)x^2+(a+b)^2x+ab(a+b)
on solving the above equation by formula -b+root over(b^2-4ac) and -b-root over(b^2-4ac)
we got -a and -b as roots.

Like? Yes (9) | No (10)  

rohit kumar   11 years AGO

{1/(a+b+x)}=(1/a)+(1/b)+(1/x)
{1/(a+b+x)}-1/x=(a+b)/ab
on equating we get
x^2+(a+b)x+ab=0
on solving the quadratic equation:
x=-a & x=-b

Like? Yes (7) | No (8)  

Jigyasa Kumar   11 years AGO

x= -a and x= -b satisfies the equation

Like? Yes (9) | No (10)  

Vikram Saini   11 years AGO

ans is option 4

Like? Yes (8) | No (9)  

Bholanath    11 years AGO

{1/(a+b+x)}=(1/a)+(1/b)+(1/x)
x^2+(a+b)x+ab=0
so, x=-a,-b

Like? Yes (9) | No (10)  

SANJIB PALUI   11 years AGO

{1/(a+b+x)}-1/x=(a+b)/ab
(x-(a+b+x))/x(a+b+x)=(a+b)/ab
-(a+b)/(x^2+x(a+b))=(a+b)/ab
x^2+x(a+b)+ab=0
x=(-(a+b)+(a-b))/2
=-b
x=(-(a+b)-(a-b))/2
=-a
[quadratic formula i.e ax^2+bx+c=0]

Like? Yes (4) | No (5)  

Maya   11 years AGO

1/(a+b+x)=(1/a)+(1/b)+(1/c)
=>1/(a+b+x)=bx+ax+ab/abx
=>abx=abx+b^2x+bx^2+a^2x+abx+ax^2+a^2b+ab^2+abx
=>ax^2+bx^2+a^2x+b^2x+2abx+a^2b+ab^2=0
=>x^2(a+b)+x(a+b)^2+ab(a+b)=0
=>x^2+x(a+b)+ab=0
=>(x+a)(x+b)=0
Therefore,x=-a,x=-b

Like? Yes (9) | No (10)  

Shashi Kumar Gupta   11 years AGO

above eqtn can be written as..
{1/(a+b+x)}-(1/x) = (1/a)+(1/b) ....
on solving above eqtn..
we get
(-x)*(a+b+x)=a*b ....

by hit n trial we get x=-a and x=-b

Like? Yes (4) | No (5)  

tirtharaj   11 years AGO

1/(a+b+x)-(1/x)=(1/a)+(1/b)
=>x-a-b-x/x(a+b+x)=(a+b)/ab
=>-1/x(a+b+x)=1/ab [as (a+b)not =0]
=>x^2+x(a+b)+ab=0
so x=-(a+b)+- sq root of ((a+b)^2-4ab))/2
so x=-a,-b

Like? Yes (3) | No (4)  

ch.jyothsna   11 years AGO

{1/(a+b+x)} = bx+ax+ab/abx
then a+b+x = abx/bx+ax+ab
then x = (abx/bx+ax+ab)-(a+b)
solving this we will get x^2+ax+bx+ab=0
x=-a x=-b

Like? Yes (4) | No (5)  

Rats   11 years AGO

when we put x=-a
{1/(a+b-a)}=(1/a)+(1/b)-(1/a)
1/b=1/b........proved
when we put x=-b
{1/(a+b-b)}=(1/a)+(1/b)-(1/b)
1/a=1/a......proved
Hence x=-a, x=-b

Like? Yes (10) | No (12)  

mounica chowdary   11 years AGO

solving the given equation we get
x^2+x*(a+b)+ab=0
x(x+a)+b(x+a)=0
(x+a)(x+b)=0
where x=-a;x=-b

Like? Yes (13) | No (15)  

Udhaya kumar   11 years AGO

from given we get x^2+x(a+b)+ab=0
then we get x=-a and x=-b..........

Like? Yes (2) | No (4)  

Parveen Soni   11 years AGO

On solving given eqn we have
(a+b) (x^2+(a+b)*x+ab)=0
since a+b is not zero hence the second on ewill be zero which result into x=-a and x=-b
option 4

Like? Yes (5) | No (8)  

Yogesh Bairagi   11 years AGO

Given equation satisfy x=-a and x=-b

Like? Yes (3) | No (6)  

D venkatesh babu   11 years AGO

sustitute in the equation..

Like? Yes (3) | No (6)  

khushboo singh   11 years AGO

becoz for both values of x i.e -a and -b
LHS=RHS

Like? Yes (6) | No (9)  

Himanshu Kumar Gupta   11 years AGO

1/(a+b+x) = 1/a+1/b+1/x
On simplyfying we get
X-a-b-x/x(a+b+x) = a+b/(ab)
We find quadratic eqn
X^2+(a+b)x+ab = 0
On solving we get x=-a and x= -b

Like? Yes (6) | No (9)  

KALIDASS   11 years AGO

1/(a+b+x)=(1/a)+(1/b)+(1/x)
1/(a+b+x)=(bx+ax+ab)/(abx)
((a+b)^2)x+(a+b)x^2+ab(a+b)=0
x^2+(a+b)x+ab=0
x=-a,x=-b

Like? Yes (8) | No (11)  

Prakash Singh   11 years AGO

solving the equation we get x= -a,-b

Like? Yes (4) | No (7)  

vikash kumar   11 years AGO

after solving equation , we get
(x+a)(x+b)(a+b)=0
thus,x=-a,x=-b

Like? Yes (3) | No (6)  

Fighter   11 years AGO

by solving eqn we get x=-a,-b

Like? Yes (2) | No (5)  

ashiwani kumar singh   11 years AGO

go to answer option choose the value of x who staisfy the condition for putting the value of x in problem
l.h.s.=r.h.s.

Like? Yes (2) | No (5)  

SUNDARAM   11 years AGO

On simplifying the given equation we get,

(x^2)+x(a+b)+ab=0

by solving the above equation we get,

x=-a,x=-b

Like? Yes (2) | No (5)  

Mayuresh Srivastava   11 years AGO

After solving given equation
x^2+x(a+b)+ab=0
(x+a)(x+b)=0
x=-a,x=-b.

Like? Yes (10) | No (14)  

rohit kumar   11 years AGO

given that,
{1/(a+b+x)}={1/a+1/b+1/x}
=>{(bx+ax+ab)/abx}=1/(a+b+x)
=>abx=(a+b+x)(bx+ab+ax)
after solving,we get
=>(x+a)(x+b)=0
=>x=-a or x=-b

Like? Yes (5) | No (9)  

MUKESH KUMAR   11 years AGO

since x=-a,x=-b satisfies the equation.

Like? Yes (7) | No (11)  

Aarush Singh   11 years AGO

x= -a and x=-b only satisfies the given equation

Like? Yes (3) | No (7)  

Bingumalla Abinay   11 years AGO

substituting we get answer

Like? Yes (3) | No (7)  

Nirmalram   11 years AGO

trial method,
substitute the given answers in the given equation
with a,b with random numbers except 0.
option 4) satisfies those condition

Like? Yes (3) | No (7)  

sugyani dalai   11 years AGO

put the value of x = -a 0r x = -b
which satisfies the equation.

therefore x=-a,x=-b

Like? Yes (4) | No (8)  

ravi singh   11 years AGO

answer is X=-a , X=-b

solve simple equation

Like? Yes (4) | No (8)