If ax2+bx+6=0 does not have two distinct real roots, where a∈R,b∈R, then the least value of 3a+b is

A
$4$
B
$- 1$
C
$1$
D
$- 2$

Solution:

It is given that

a x^{2} + bx + 6 = 0

does not have two distinct real roots.
Discriminant is given by

D = b^{2} - 4 ac

.
Here are the relations between roots and discriminant
●       When roots are non-real, the discriminant is less than 0
●       When roots are real and equal, the discriminant is equal to 0
●       When the roots real and unequal the discriminant is greater than 0
We substitute

b = k - 3 a

to obtain

a x^{2} + (k - 3 a) x + 6 = 0

.
Since, the equation does not have real distinct roots.
Therefore,

D = {(k - 3 a)}^{2} - 24 a \leq 0

\Rightarrow 9 a^{2} - 6 a (4 + k) + k^{2} \leq 0

Above quadratic equation has real roots.
Therefore,

D = 36 [{(4 + k)}^{2} - k^{2}] \geq 0

\Rightarrow k \geq - 2

\Rightarrow 3 a + b \geq - 2

Hence, the correct option is 4.

If $a x^{2} + bx + 6 = 0$ does not have two distinct real roots, where $a \in R, b \in R$ , then the least value of $3 a + b$ is

Solution:

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