1. ## Quadratics? Simultaneous equations, not understanding how to finish problem.

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

$\displaystyle c^2-12c-124>0$
$\displaystyle -6.65<c<18.65$

2. 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

$\displaystyle c^2-12c-124>0$
$\displaystyle -6.65<c<18.65$
should be $\displaystyle c^2-12c-124<0$

3. ah, ok, oops. weird mistake, sorry.

in that case, i understand how you get

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

but i still don't understand how to get
$\displaystyle -6.65<c<18.65$

4. Originally Posted by ns2583
ah, ok, oops. weird mistake, sorry.

in that case, i understand how you get

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

but i still don't understand how to get
$\displaystyle -6.65<c<18.65$
counting b^2-4ac,then discuss...

5. Originally Posted by repcvt
counting b^2-4ac,then discuss...
• Solve x2 + 2x – 8 < 0.

• First, I'll find the zeroes:
• x2 + 2x – 8 = 0
(x + 4)(x – 2) = 0

x = –4
or x = 2
These x-intercepts split the number line into three intervals: x < –4, –4 < x < 2, and x > 2. Since this is a "less than" inequality, I need the intervals where the parabola is below the x-axis. Since the graph of y = x2 + 2x – 8 is a right-side-up parabola, it is below the axis in the middle (between the two intercepts). Since this is an "or equal to" inequality, the boundary points of the intervals (the intercepts themselves) are included in the solution.
Then the solution is: –4 < x < 2
You can check the answer from the graph:

more Solving Quadratic Inequalities: More Examples