# 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.

Q:

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:

$$[0,-\frac{b}{a}]$$

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.

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

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

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

• 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 Sep 10, 2021 at 0:53
• @Vega did you find it at the end? Sep 10, 2021 at 10:23
• Sorry, I still cannot figure it out :( Sep 10, 2021 at 11:10
• see now, I updated! Sep 10, 2021 at 13:19
• Thanks a lot, I understood it. 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 $$\quad x=y=0\qquad a,b\in\mathbb{Z^+}\\ \quad x=y=-1\quad a=2,\space b=1\\ \quad x=y=-2\quad a=2,\space b=2\\ \quad x=y=-3\quad a=2,\space b=3\\ \quad x=y=-4\quad a=2,\space b=4\\ \vdots \\ \quad x=y=-n\quad a=2,\space b=n$$