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

Find the values of k so that the equation x22x(1+3k)+7(3+2k)=0 has real and equal roots.

Open in App
Solution

The given equation is x22x(1+3k)+7(3+2k)=0.

It has real and equal roots. So, b24ac=0.

[2(1+3k)]24(1)(7)(3+2k)=0

4(1+9k2+6k)4(21+14k)=0

4+36k2+24k8456k=0

36k232k80=0

9k28k20=0

9k218k+10k20=0

9k(k2)+10(k2)=0

(k2)(9k+10)=0

k=2,109

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