# Thread: Help me on a question

1. ## Help me on a question

Question 5 AEA summer 2002:

For part (a), I did $\displaystyle -kx(2+x)=(2.5)^2-x^2$, and I got $\displaystyle (1-k)x^2-2kx-(2.5)^2=0$. I then used the quadratic formula to get $\displaystyle 4k^2-25k+25=0$, giving $\displaystyle k=5$ or $\displaystyle \frac{5}{4}$ (because if $\displaystyle f$ touches $\displaystyle g$, then $\displaystyle b^2=4ac$). However, the next step I'm confused about. Please help. What value of $\displaystyle k$ do I choose?

2. Hello Rednest
Originally Posted by rednest
However, the next step I'm confused about. Please help. What value of $\displaystyle k$ do I choose?
The equation $\displaystyle -kx(2+x)=2.5^2-x^2$ is the same as $\displaystyle f(x)=g(x)$ only if $\displaystyle x$ lies in $\displaystyle [-2,0[$. If you solve the two equations $\displaystyle -kx(2+x)=2.5^2-x^2$ for $\displaystyle x$ with $\displaystyle k=5$ and $\displaystyle k=5/4$ you'll get a solution $\displaystyle x\in[-2,0[$ and another one $\displaystyle x\not\in[-2,0[$ : the right value of $\displaystyle k$ is the one such that $\displaystyle x\in[-2,0[$.