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

If 4x6+ax3+5x7=0 has a maximum of 4 possible real roots, then which of the following is true?

A
a<0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
a=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
None of the above
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
a>0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C None of the above
Given that f(x)=4x6+ax3+5x7=0

According to Descartes’ Rule of Signs the number of positive real zeros is either equal to the number of sign changes of f(x) or is less than the number of sign changes by an even integer and the number of negative real zeros is either equal to the number of sign changes of f(x) or is less than the number of sign changes by an even integer.
Let a>0,
then in the given f(x) we have one sign change, so there can be one positive real root.
f(x)=4x6|a|x35x7
In the given f(x) we have one sign change, so there can be one negative real root.
Therefore the total real roots can be 1+1=2
But it is given that there are 4 possible real roots.

So, now let's take a<0,
then in the given f(x) we have three sign changes, so there can be three positive real roots or one positive real root.
Also, f(x)=4x6(a)x35x7
f(x)=4x6+ax35x7
Since a<0, the given f(x) will have one sign change, so there can be one negative real root.
Therefore the total real roots can be 4 or 2
Thus, no real values of a is possible.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Descarte's Rule for Positive Roots
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon