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

Let a>b>0 and the quadratic equation ax2+bx9=0 has roots as α,β (α<β). If a3+b3+27ab=729, then the value of 4βaα is

Open in App
Solution

Given quadratic equation is ax2+bx9=0
a3+b3+27ab=729a3+b3729+27ab=0a3+b3+(9)3=3ab(9)
a+b9=0 or a=b=9
As a>b>0, so a+b9=0

Now, f(x)=ax2+bx9
Putting x=1, we get
f(1)=a+b9=0
So the quadratic equation has one root as 1
Other root =9a
Therefore, α=9a, β=1
4βaα=4+9=13

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Relation of Roots and Coefficients
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon