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

If (2p+1)x2(p+2)x+p1=0 has repeated roots, then 7p28p=

Open in App
Solution

Condition for repeated roots
b2=4ac

(p+2)2=4(2p+1)(p1)

p2+4p+4=8p28p+4p4

p2+4p+4=8p24p4

7p28p8=0

7p28p=8

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Zeroes of a Polynomial concept
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon