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

The roots of the equation, 6x425x3+12x2+25x+6=0 are

A
32,2,23,3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
12,2,13,3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
12,2,13,3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
32,2,23,3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 12,2,13,3
Given; 6x425x3+12x2+25x+6=0
Dividing the equation by x2, we get:
6(x2+1x2)25(x1x)+12=0
Let: x1x=t
x21=xt
x2xt1=0
Now, this equation should have real roots:
D0t2+40, which is true tR
Now, Squaring both sides, we get:
x2+1x2=t2+2
Substituing both we get the above equation as:
6t225t+24=0
6t29t16t+24=0
3t(2t3)8(2t3)=0
(3t8)(2t3)=0
t=83,32

If t=83, we have
x1x=83

3x(x3)+1(x3)=0
(x3)(3x+1)=0
x=13,3

If t=32, we have
x1x=32
2(x21)=3x
2x223x=0
2x24x+x2=0
(2x+1)(x2)=0
x=12,2

All possible values of x are 12,2,13,3






flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Biquadratic Equations of the form: ax^4+bx^3+cx^2+bx+a=0
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon