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If the equation ax^2 +bx + 6 = 0 has real roots where a b are elements of R then the greatest value of 3a+b is a)4 b)-1 c)-2 d) 1

```
2 years ago

Vasudeva Bhat
14 Points
```							Take x=3,=>f(3)≥0=>9a+3b+6≥0=>3a+b≥-2Hence, we can get 3a+b=-2 as the minimum value.    Thanks    Cheers.
```
2 years ago
Meghendra Agrawal
8 Points
```							Please tell why can’t f(3) Please give the answer soon. or
```
2 years ago
Meghendra Agrawal
8 Points
```							Why can’t f(3) be less than 0?? Can we also assume x = 1 or 2 or – 1 but how will we determine whether f(x) would be greater than or less than 0??
```
2 years ago
Vasudeva Bhat
14 Points
```							I checked with the notes my sir had provided and I realised the question you've given needs to be corrected. In the actual question, there are no real roots. So The line won't intersect x axis at all that means ax^2+bx+c>0We substitute x=3 because we need to get 3a+b and one of the ways we can obtain it is by substituting 3 and we get 3a+b>-2I hope your doubt would have been cleared by now. Thanks Cheers!
```
2 years ago
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### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions