Curve and line

The equation of a curve C is y=2x² -8x+9 and the equation of a line L is x+ y=3

(1) Find the x co-ordinates of the points of intersection of L and C.
(2) Show that one of these points is also the stationary point of C?

Result

a1 =  2
a2 =  1.5
b =  2

Solution:

 3−x=2⋅ x2−8⋅ x+9 −2x2+7x−6=0 2x2−7x+6=0  a=2;b=−7;c=6 D=b2−4ac=72−4⋅2⋅6=1 D>0  x1,2=−b±D2a=7±14 x1,2=7±14 x1,2=1.75±0.25 x1=2 x2=1.5   Factored form of the equation:  2(x−2)(x−1.5)=0 a1=x1=2 \ \\ 3-x=2 \cdot \ x^2-8 \cdot \ x+9 \ \\ -2x^2 +7x -6=0 \ \\ 2x^2 -7x +6=0 \ \\ \ \\ a=2; b=-7; c=6 \ \\ D=b^2 - 4ac=7^2 - 4\cdot 2 \cdot 6=1 \ \\ D>0 \ \\ \ \\ x_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 7 \pm \sqrt{ 1 } }{ 4 } \ \\ x_{1,2}=\dfrac{ 7 \pm 1 }{ 4 } \ \\ x_{1,2}=1.75 \pm 0.25 \ \\ x_{1}=2 \ \\ x_{2}=1.5 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 2 (x -2) (x -1.5)=0 \ \\ a_{1}=x_{1}=2

Checkout calculation with our calculator of quadratic equations.

a2=x2=1.5=32a_{2}=x_{2}=1.5=\dfrac{ 3 }{ 2 }
4b−8=0 4b=8 b=8/4=2 =24b-8=0 \ \\ 4b=8 \ \\ b=8 / 4=2 \ \\ =2

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