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

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The curve y = x^2 + 3x + 2 and the line y = 3 + 2x have common points if the equations have real solutions.

Substitute y = 3 + 2x in y = x^2 + 3x + 2

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

=> x^2 + x - 1 = 0

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

=> x1 = -1/2 + (sqrt 5)/2

x2 = -1/2 - (sqrt 5)/2

The points of intersection are (-1/2 + (sqrt 5)/2, 2 + sqrt 5) and (-1/2 - (sqrt 5)/2 , 2 - sqrt 5)

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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).

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