Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
x 3 / x 3 +27+9x(x+3) = x-3 / x+6 x3/ x3+27+9x(x+3) = x-3 / x+6
x3/ x3+27+9x(x+3) = x-3 / x+6
We are given that x3/(x+3)3 = x-3/(x+6). From componendo-dividendo, we get (x+3)3-x3/(x+3)3+x3 = (x+6) - (x-3)/(x+6) + (x-3) or 3 [x2+(x+3)2 + x(x+3)]/(2x+3)(x2+(x+3)2-x(x+3)] = 9/(2x+3) So, either x = -3/2 or 3[x2+(x+3)2-x(x+3)] = x2+(x+3)2 + x(x+3) The second equation would imply 9 = 0 and hence x = -3/2 is the only possibility
We are given that x3/(x+3)3 = x-3/(x+6). From componendo-dividendo, we get
(x+3)3-x3/(x+3)3+x3 = (x+6) - (x-3)/(x+6) + (x-3)
or 3 [x2+(x+3)2 + x(x+3)]/(2x+3)(x2+(x+3)2-x(x+3)] = 9/(2x+3)
So, either x = -3/2 or 3[x2+(x+3)2-x(x+3)] = x2+(x+3)2 + x(x+3)
The second equation would imply 9 = 0 and hence x = -3/2 is the only possibility
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -