wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Evaluate 12abx2(9a28b2)x6ab=0, where a0 and b0.

Open in App
Solution

12abx2(9a28b2)x6ab=0a0,b0
Therefore, the roots are:
x=(9a28b2)±(9a28b2)2+24ab(12ab)2×12ab=(9a28b2)±81a4+64b4144a2b2+288a2b224ab
=(9a28b2)±(9a2+8b2)224ab
=(9a28b2)±(9a2+8b2)24ab
x=9a2×224ab=3a4bor,x=8b2×224ab=2b3a

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Solving QE by Completing the Square
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon