CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The equation x45x3+kx23x1=0 has

A
4 real and distinct roots
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Two real and two imaginary roots
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4 imaginary roots
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Two repeated real roots and two imaginary roots
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 4 real and distinct roots
α,β,γ,δ x45x3+kx23x1=0
α,β x2ax1=0
γδ x2bx+1=0
α+β+γ+δ=5 a+b=5
α+β=+a αβ+βγ+γδ+δα=k
γ+δ=b 1+1+βγ+δα=k
αβγδ=1
αβ=1
γδ=1
Hence sthey are satisfying αβγδ=1
4 real & distinct roots.

1202823_827604_ans_b6d8ab79fb7541eda2a03991b4b4b945.jpg

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Solving QE by Factorisation
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon