wiz-icon
MyQuestionIcon
MyQuestionIcon
5
You visited us 5 times! Enjoying our articles? Unlock Full Access!
Question

The maximum possible number of real roots of equation x56x24x+5=0 is

A
3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 3
Let f(x)=x56x24x+5=0
Then the number of change of sign in f(x) is 2 therefore f(x) can have at most two positive real roots .

Now, f(x)=x56x4+4x+5=0
Then the number of change of sign is 1.
Hence f(x) can have at most one negative real root.

So, maximum possible number of real roots is 3.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Fundamental Theorem of Algebra
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon