Question

Question in comments .

Answer 1

If \(\large \bf a,b \in R\) and \(\large \bf ax^2+bx+6=0,a \neq 0 \) does not have two distinct real roots , then
Find:-
1) the minimum possible value of 3a+b
2) the minimum possible value of 6a+b

Answer 2

\[b^2 < 24a\]Now for inequalities, you always, always, always remember that conditions like minimum hold at boundaries.

Answer 3

yup, i know

Answer 4

then

Answer 5

Alright. So we predict that this holds if and only if \(b^2 = 24 a\).
Thus\[3a+b = \frac{b^2}{8}+b\]

Answer 6

maine bhi yahi kiya tha

Answer 7

Now \(b^2/8 + b\) is a quadratic expression and you can find its minimum.

Answer 8

oh ! damn...
silly mistake !

Answer 9

multiply galat kiya

Answer 10

oh lol

Answer 11

8*1=1

Answer 12

lol

Answer 13

by the way, thanks yaar :) @ParthKohli

Answer 14

no problem

Answer 15

btw this expression will never actually reach that value

Answer 16

to kaunsi hogi

Answer 17

I mean the condition is \(b^2 < 24a\) and we've actually taken \(b^2 = 24a\) which is not possible, so \(3a + b\) will never attain its "minimum" value we calculated.

Answer 18

Actually the condition is \(b^2 \le 24a\) lol.
I skipped the part about two distinct roots.

Answer 19

Yes it will attain that value.

Answer 20

you forgot this condition

Answer 21

What condition?

Answer 22

b^2=4ac