A curve C is given by the equation y^3 + y^2 + y = X^2 - 2x. Show that the point (3,1) is the only point of intersection of the line x = 3 and the curve. Thanks!
You should start by substituting 3 for x in the equation of the curve such that:
`y^3 + y^2 + y = 3^2 - 2*3 `
`y^3 + y^2 + y = 9 - 6 `
`y^3 + y^2 + y = 3`
Notice that if you substitute 1 for y yields:
`1^3 + 1^2 + 1 = 3 =gt 3= 3`
You need to prove that the y=1 is the only solution to the equation `y^3 + y^2 + y- 3= 0`
Since y = 1 is a solution, you may use this to write the factored form of equation such that:
`y^3 + y^2 + y - 3 = (y-1)(ay^2 + by + c)`
You should identify a,b,c such that:
`y^3 + y^2 + y - 3 = ay^3 + by^2 + cy - ay^2- by- c`
Equating the coefficients of like powers yields:
`-c = -3 =gt c = 3`
`c - b = 1 =gt3 - b= 1 =gt -b=1 - 3 =gt b = 2`
`ay^2 + by + c = y^2 + 2y + 3`
You need to find the roots of equation`y^2 + 2y + 3 = 0` to verify if the solution y=1 is unique such that:
`y_(1,2) = (-2+-sqrt(4-12))/2`
Notice that `sqrt(4-12) = sqrt(-8) !in R` , hence, the equation y`^2 + 2y + 3 = 0` has no real solutions.
This proves that y=1 is the only real solution to equation `y^3 + y^2 + y - 3 = 0.`
Hence, the curve `y^3 + y^2 + y = x^2 - 2x ` only intersects the line x=2 at the point (3,1).