MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price:

There are no items in this cart.
Continue Shopping
Menu
Vignesh A.S. Grade: 8
        

Find the square root of 1+ ((8x^3 + 8x) / (x^4 - 4x^3 + 5x^2 - 4x +4))

6 years ago

Answers : (2)

Priyansh Bajaj AskiitiansExpert-IITD
30 Points
										

Dear Vignesh,

Solution:- x4 -4x3 +5x2 -4x+4 = x4 + x2 -4x3 -4x +4x2 +4


                                        = x2 (x2 +1) -4x(x2 +1) +4(x2 +1)


                                        = (x2 +1)(x2 -4x +4)


                                       = (x2 +1)(x -2)2


And, 8x3 + 8x = 8x(x2 +1)


To find sq. root of {1+ [8x(x2 +1)] / [(x2 +1)(x -2)2]}


or {1+ 8x/ (x -2)2}1/2 = {(x +2)2/ (x -2)2}1/2


                              = (x +2)/ (x -2) [ANS]



Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.


All the best Vignesh!!!



Regards,

Askiitians Experts

Priyansh Bajaj

6 years ago
akash kumar
16 Points
										

     =1+(8x^3)/(x^4-4x^3+5x^2-4x+4


     =(x^4-4x^3+5x^2-4x+4+8x^3+8x)/(x^4+4x^3+5x^2-4x+4


     =-1



6 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: 15,000
  • View Details
Get extra R 3,000 off
USE CODE: CART20
Get extra R 440 off
USE CODE: CART20

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details