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

If f(x)=4x4+12x3+cx2+6x+d is a perfect square, then

A
c=5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
c=13
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
d=1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Minimum value of f(x) occurs at x=12
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D Minimum value of f(x) occurs at x=12
f(x)=4x4+12x3+cx2+6x+d
If a bi-quadratic polynomial is a perfect square, then it must be in the form of
(ax2+px+q)2
=a2x4+2apx3+(p2+2aq)x2+2pqx+q2
Comparing the coefficient, we get
a2=4a=±2
2ap=12p=±3
2pq=6q=±1
a,p,q have same sign.

Now, d=q2d=1
c=p2+2aq
=9+2(±2)(±1)
=9+4
=13

f(x)=(2x2+3x+1)2
=4x4+12x3+13x2+6x+1
The minimum value of f(x) is 0
(2x2+3x+1)2=0
2x2+3x+1=0
(2x+1)(x+1)=0
x=12 or 1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Playing With It Part 2
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon