Tuesday, July 2, 2013

How to verify if the curve y=x^2+3x+2 and the line y=3+2x have common points?

If the given curve and line have common points, then the
system formed from their equations have solutions. The solutions of the system represent
the intercepting points of the line and the curve.


We'll
equate x^2+3x+2 and
3+2x


x^2+3x+2=3+2x


We'll shift
all terms to one side and we'll combine like terms:


x^2 +
3x + 2 - 2x - 3 = 0


x^2 + x - 1 =
0


We'll apply quadratic
formula:


x1 = [-1+sqrt(1 +
4)]/2


x1 = (-1+sqrt5)/2


x2 =
(-1-sqrt5)/2



Now, we'll substitute the value of
x in the equation of the line, because it is much more easier to compute
y.


y1=2*(-1+sqrt5)/2+3


y1 = 2
+ sqrt5


The first intercepting point is: A((-1+sqrt5)/2 ; 2
+ sqrt5)


y2 =
2*(-1-sqrt5)/2+3


y2 = 2 -
sqrt5


The  2nd intercepting point is: B((-1-sqrt5)/2 ; 2 -
sqrt5)


Therefore, the common points are:
A((-1+sqrt5)/2 ; 2 + sqrt5) and B((-1-sqrt5)/2 ; 2 -
sqrt5).

No comments:

Post a Comment