# Finding value of k for which fg(x)=k has equal roots?

I've been going through this community and I Find this really helpful. About me(I know I should be precise but ya), I'm just a highschool student who can't afford any coachings/schools. Self schooling being my only option I'm trying to teach myself mathematics from some torrented books. I am on functions and their graphs and stuck with one question

The functions f and g are defined for x ∈ R by f(x) =4x − 2x^2; g(x)= 5x + 3.
(i)  Find the range of f. (ii)  Find the value of the constant k for which the equation gf(x) = k has equal roots.

Now, I do understand the composition of functions but I just don't understand what they are asking in this case, g(fx()=k would result in

20x-10x^2 +3=k

Now this is a quadratic equation of second degree which should have 2 roots/solutions, but that's the case if the right hand side was zero and not k. I have absolutely no idea how to tackle this question and what's being asked in part ii of the question. Anyhelp would be highly appreciated

• Notation $gf(x)=k$ is ambiguous. Is it $g(f(x))=k$ or $g(x)f(x)=k$? Aug 27, 2016 at 8:11
• Hi, I see this is about three years old now, but I wanted to point out that you should accept answers you find useful. I was surprised it popped up in the feed even though it appears no activity has occurred here recently. Apr 25, 2019 at 7:29

Your equation could be rewritten as $$10x^2-20x+(k-3)=0.$$

Recall that the roots of a quadratic equation $ax^2+bx+c=0$ are given by $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ So the two roots are equal when $b^2-4ac=0$. Can you apply this to your problem?

• Hey, Thank you for the answer! I really appreciate that. Now back to the question b^2 -4ac=0 would result in k=10.3 (I Just isolated the k) There are a number of questions I have here, I am equally embarrassed of my ignorance but here it goes How did you know that the two roots would be equal when b^2-4ac=0 ? The two roots are equal when the equation is a perfect square since (x+2)^2 can be written as (x+2)(x+2), that's just my intuition but how did you know that b^2-4ac =0 is a case for that too? Secondly, I thought(and I know I've been wrong) that they were asking a value (cont.) Aug 27, 2016 at 6:59
• (continued) a value of k for which there would be k roots, my bad Aug 27, 2016 at 6:59
• Your $k=10.3$ is not quite correct. You're solving $20^2-4\cdot 10\cdot (k-3)=0$.
– pi66
Aug 27, 2016 at 7:01
• If you look at the root formula that I wrote, you can see that the two roots are equal only if the term $\sqrt{b^2-4ac}$ is zero. This corresponds to the case where you can write the polynomial as $(x+d)(x+d)$, where $d$ is the root that appears twice.
– pi66
Aug 27, 2016 at 7:02
• Oops, my bad! k=13. I see, since adding and subtracting zero would be the same. Thank you for this. That was really helpful :) Aug 27, 2016 at 7:15