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

Calculate 1x31+1x32, where x1 and x2 are roots of the equation 2x23ax2=0.

Open in App
Solution

2x23ax2=0
Since x1,x2 are the roots of the given equation
x1+x2=(3a/2)=3a/2
x1x2=2/2=1
1x31+1x32=x31+x32x21x32=x1+x2(x1+x2x1x2)x31x32
Now, x21+x22=(x1+x2)22x1+x2=(9a24+2)
1x31+1x32=x1+x2(x1+x2x1x2)x31x32=3a2(9a24+2)=3a(9a2+8)8

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
AM and GM
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon