I'm struck on this question, I tried hard but couldn't solve it.

Question: if a quadratic equation in $x$: $$ax^2 - bx + 5 = 0$$ does not have two distinct real roots, then find the minimum value of $5a + b$.

So far, I tried using the condition that the discriminant should be negative or zero, but couldn't proceed further.

Moreover, as the given equation doesn't have two distinct roots so the graph will be either concave upwards or downwards, by double differentiation, I found that graph will be concave upwards so this equation will be positive or zero for all real $x.$

Any help will be appreciated.


$$D\leq0$$ $$\Longrightarrow b^2\leq20a$$ $$\Longrightarrow 5a\geq\frac{b^2}{4}$$ $$\Longrightarrow 5a+b\geq \frac{b^2}{4}+b\geq -1$$ where equality holds if $b = -2\ and\ a = \frac{1}{5}$

Also, you need not double differentiate to get that the graph is concave upwards, just see that it is taking positive value at 0, so it will take non-negative values for all real $x$, hence, it must be concave upwards.

Hope it is helpful:)

  • 1
    $\begingroup$ Good answer! Just to make things clearer, that last $\frac{b^2}{4}+b\geq -1$ can be easily checked by completing the square. $\endgroup$ Dec 8 '18 at 12:59
  • 1
    $\begingroup$ Yes, that is how I did it. $\endgroup$
    – Martund
    Dec 8 '18 at 13:01
  • $\begingroup$ Thanks for help..finally I know how to deal with these problems. $\endgroup$
    – Shivansh J
    Dec 8 '18 at 16:03

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