# Math Help - Help me on a question

1. ## Help me on a question

Question 5 AEA summer 2002:

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

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