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If $ax^2 - bx + 5 =0$ does not have two distinct real roots, what is the minimum value of $5a + b$?
I tried taking discriminant $< 0$ to get,
$b^2 \le 20a \implies -2\sqrt{5a} \le b \le 2\sqrt{5a}$
That's about it. I'm not able to proceed any further... can somebody please help?
We have that
$$b^2-20a\le 0 \implies a\ge \frac{b^2}{20}$$
thus
$$5a+b\ge f(b)=\frac {b^2}{4}+b$$
$$f'(b)=\frac{b}2+1=0\implies b=-2$$
$$f''(b)=\frac12>0$$ $$\implies \min(5a+b)=-1$$
correct me if wrong:
Should be $b^2\le 20a$, implies:
$-2\sqrt{5a}\le b$ implies :
$5a -2\sqrt{5a} \le 5a+b.$
$(\sqrt{5a} -1)^2 -1 \le 5a+b.$
LHS we got a square $ \ge 0.$
The minimum value for $5a+b$ is?
You have $a \ge \dfrac {b^2} {20}.$ The graph of that is bounded by the parabola $a = \dfrac {b^2} {20}$ in the $(a,b)$-plane. Subject to that constraint you want to minimize $5a+b.$ Making the $b$-axis horizontal and the $a$-axis vertical, the region is above the parabola. Draw the parabola and the line whose equation is $5a+b=\text{constant}$ and you see that as that as the "constant" gets smaller, the line moves toward a position where it just touches the parabola at one point. That one point is where the slopes of the line and the parabola are equal. The slope of the line is $-1/5$ since the equation is $a=\frac{-1} 5 b+\text{constant}.$ So you need the point on the parabola where the slope is $-1/5.$ The slope is given by $$ \frac{da}{db} = \frac d {db}\, \frac{b^2}{20} = \frac b {10}. $$ This is equal to $-1/5$ when $b=-2.$ So the point $(b,a) = (-2, 1/5).$ So $5a+b = -1.$
We can obtain the result with some geometry:
The equation has not two distinct real roots if and only if its discriminant $b^2-20a\le 0$, i.e. if and only if, in the $(a,b)$-plane, the point $(a,b)$ is inside the parabola $\mathcal P$ with equation $b^2=20a$.
Consider the family of straight lines with equations $\;5a+b=k\enspace(k\in\mathbf R)$. The minimum sought for is the value of $k$ for which the corresponding straight line has a double intersection with $\mathcal P$.
This let's determine the intersection points: they satisfy the equations $b^2=20a\;$ and $\;5a=k-b$, whence $$b^2=4(k-b)\iff b^2+4b=4k\iff(b+2)^2=4(k+1).$$ We have a double intersection if and only if $b=-2$, which corresponds to $\color{red}{k=-1}$.
First you need to replace your "$<$" by "$\leq$".
Then, in $-2\sqrt{5a} \leq b \leq 2\sqrt{5a}$ you can add $5a$ everywhere...
