If $b>0$ then find the number of values of'a' for which domain and range of $f\left(x\right)=\sqrt{ax^{2}+bx}$ are equal.

My approach:

Since we have a square root, $$ax^{2}+bx\ge0\to x\left(ax+b\right)\ge0$$ If $a>0$ then domain is: $$x \in \ \left(-\infty,-\frac{b}{a}\right] \cup \ \left[0,\infty\right)$$

If $a<0$ then domain is:


If $a=0$ then domain is: $$x\ge0$$

But I don't know how to proceed further with the main question. Any hints or suggestions are welcome.


2 values of a are possible.

  • $\begingroup$ The $a$ values that satisfy this condition are $-4$ and $0$. Graphing both of these functions should give you an idea why. $\endgroup$ Sep 9, 2021 at 17:44
  • $\begingroup$ To each other! That's a common construction in English. $\endgroup$
    – user247327
    Sep 9, 2021 at 19:44

2 Answers 2


If $a < 0$ the domain you pointed out is wrong , the right one is $[0,-\frac{b}{a}]$.

One value is $a = 0$ because then $f(x) = \sqrt{bx}$ which has range $[0,\infty]$.

For $a > 0$ there can't be no value having domain = range because your domain has always negative values inside while range, because of the presence of the square root, needs to contain just positive values.

So the second value has $a < 0$, now you can see that $f(0) = f(-\frac{b}{a}) = 0$ and that is the minimum value of the range , consequently the maximum is attained inside the interval so you can just use a derivative test to find which one is it :

$$f'(x) = 0 \implies \frac{2ax+b}{2 \sqrt{ax^2+bx}} = 0 \implies x = -\frac{b}{2a}$$

and so the maximum is $f(-\frac{b}{2a}) = b\sqrt{\frac{1}{4a}-\frac{1}{2a}}$, but then to have domain = range you need that

$$b\sqrt{\frac{1}{4a}-\frac{1}{2a}} = -\frac{b}{a} \implies a =-4$$

  • $\begingroup$ So you mean for a>0, there are still negative values in domain while range is positive. I got this part. But I still couldn't find how to get -4 $\endgroup$ Sep 10, 2021 at 0:53
  • $\begingroup$ @Vega did you find it at the end? $\endgroup$
    – Tortar
    Sep 10, 2021 at 10:23
  • $\begingroup$ Sorry, I still cannot figure it out :( $\endgroup$ Sep 10, 2021 at 11:10
  • $\begingroup$ see now, I updated! $\endgroup$
    – Tortar
    Sep 10, 2021 at 13:19
  • $\begingroup$ Thanks a lot, I understood it. $\endgroup$ Sep 10, 2021 at 14:10

I interpret your question as wanting to find if and when $\space x=y.\quad $ I thought I had it immediately until I noticed the rquirement: $\space b>0$ $$ f(x)=y=\sqrt{ax^{2}+bx}\\ ax^{2}+bx-y^2=0\\ $$ by inspection we can see first solutions and then generalize the pattern $$ \quad x=y=0\qquad a,b\in\mathbb{Z^+}\\ \quad x=y=-1\quad a=2,\space b=1\\ \quad x=y=-2\quad a=2,\space b=2\\ \quad x=y=-3\quad a=2,\space b=3\\ \quad x=y=-4\quad a=2,\space b=4\\ \vdots \\ \quad x=y=-n\quad a=2,\space b=n$$


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