Originally Posted by

**ns2583** Okay. Here goes... Thanks in advance to anyone who can help.

Find the range of values of *c* for which the straight line y=3x+c intersects the curve $\displaystyle x^2+4x+y^2=12$ in 2 distinct points

so this is how i did it (and it was all right, according to the answers- ill bold the part that i don't get why you do)

$\displaystyle y=3x+c$

$\displaystyle x^2+4x+y^2=12$

$\displaystyle x^2+4x+(3x+c)^2=12$

$\displaystyle x^2+4x+9x^2+C^2=12$

$\displaystyle 10x^2+4x+6cx+c^2-12=0$

$\displaystyle 10x^2+(4+6c)x+C^2-12=0$

$\displaystyle (4+6c)^2-4*10*(C^2-12)>0 $(because 2 distinct points)

$\displaystyle 16+48c+26c^2-40c^2+480>0$

$\displaystyle -4c+48c+496>0$

$\displaystyle 4(-1+12c+124)>0$

**OK, AND THIS IS THE LAST PART, BUT I HAVE NO CLUE HOW( OR WHY ITS DONE) TO GET THIS**

$\displaystyle c^2-12c-124>0$

$\displaystyle -6.65<c<18.65$