plz solve: simultaneous equation?

Question

find the point(s) of intersection betweeen the parabola y=x^2+3x+2 and the line y-7x+2=.0

1)ow many point(s) of intersection are there between between the line and the curve?

Answer 1

To find the point of intersection between two curves, put y1=y2 so

Let
y1 = x^2 + 3x +2
y2 = 7x -2

y1 = y2

x^2 + 3x + 2 = 7x -2

Rearrange:

x^2 + 3x + 2 - 7x + 2 = 0 ---> x^2 - 4x + 4 =0

This is a second order eq. to solve it

x1,x2 = [-b +- Sqr(b^2 -4 a c )] / (2 a)

where a = 1, b = -4 and c = 4

x1,x2 = [+4 +- Sqr( (-4)^2 - 4 (1) (4) ) ] / (2*1)

x1,x2 = [+4 +- Sqr( 16 - 16) ] / 2

x1,x2 = 4/2 = 2

then x1=x2= 2 that means there is only one point; to find the corresponding y substitute x in either of the eq.

y = 7 (2) -2 = 12

Then the point (2,12) is the point of tangency because a parabola and a line intersect with two point. Therefore, (2,12) is the tangency point..!

Answer 2

There are 2 points of intersection (0, 2) and (3, 30)

Answer 3

I agree with the first answer - intersection at (2,14)

To solve, solve the second equation for y and equate the two expressions for y:

x^2 + 3x + 2 = 7x - 2

which simplifies to

x^2 - 4x + 4 = 0

whose two roots are the same, namely, x=2

whence y = 12

Answer 4

y = x^2 + 3x + 2
y - 7x + 2 = 0

y - 7x + 2 = 0
y = 7x - 2

7x - 2 = x^2 + 3x + 2
x^2 - 4x + 4 = 0
(x - 2)(x - 2) = 0
x = 2

y = 7x - 2
y = 7(2) - 2
y = 14 - 2
y = 12

ANS : (2,12), there is only 1 point between the line and curve.

Answer 5

They intersect only once, at (2,12).