$ ax^2 + bx +6 $ does not have two distinct real roots , then what will be the least value of $ 3a + b $?

Question

$ax^2 + bx +6$ does not have two distinct real roots , then what will be the least value of $3a + b$?

I know that $D$ will be less than or equal to. But least value of $3a + b$ can not be deduced from that.

Can anyone please help me?

Answer 1

Guide:

We want to minimize $3a+b$ subject to $b^2-24a \le 0$.

When the minimal is attained, $b^2=24a$ is satisfied. That is $a=\frac{b^2}{24}$.

Hence, we want to minimize $\frac{3b^2}{24}+b=\frac{b^2}{8}+b$, can you take it on from here? Say, by using calculus or completing the square.

Edit:

Relevant Desmos link. As we change the value of $k$, notice that the optimal value must touches the boundary.

Answer 2

The solutions to a quadratic are given by

$$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$$

In order for a quadratic equation to have 2 distinct real roots, $\sqrt{B^2 - 4AC}$ must product 2 distinct real values. Which only happens when $B^2 - 4AC > 0$. So you want $B^2 - 4AC \le 0$.

See how far you can work it out with that.

Answer 3

Solution

If $a=0$ and $b \neq 0$, the equation degenerates into $bx+6=0(b \neq 0)$. Since any $b \neq 0$ may satisfy the requirement we raise, hence there exists no minimum value for $3a+b$, namely, $b$.

Now, let's consider the situation when $a \neq 0.$ In this case, we need and only need that $$\Delta=b^2-24a \leq 0.\tag 1$$

Denote $3a+b=k$. Then $b=k-3a$.Putting it into $(1)$, we obtain that $$(k-3a)^2-24a=9a^2-(6k+24)a+k^2 \leq 0, \tag2$$ which could be regarded as a one-variable quadratic inequality with respect to $a.$ Moreover, this shows that there exists at least one solution for the inequality $(2)$. Thus, we may claim $$\Delta'=[-(6k+24)]^2-4 \cdot 9 \cdot k^2=288(k+2) \geq 0.$$ Therefore, $$k \geq -2.$$ As a result, the mimimum value of $3a+b$ is $-2.$

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Note

From the geometrical view, we may see that all the lines on and above $3a+b=-2$ could satisfy the requirement.

Answer 4

Let $f(x)=ax^2+bx+6$ and $f(0)=6>0$

So function $f(x)\geq 0$ for all real $x$

So $f(3)=9a+3b+6\geq0\Rightarrow 3a+b\geq -2$