Distance problem

A=(x, x)
Distance AB=√5, find x;


x1 =  3
x2 =  2


(x1)2+(x4)2=5   2x210x+12=0  a=2;b=10;c=12 D=b24ac=1024212=4 D>0  x1,2=b±D2a=10±44 x1,2=10±24 x1,2=2.5±0.5 x1=3 x2=2   Factored form of the equation:  2(x3)(x2)=0 x1=3(x-1)^2 + (x-4)^2 = 5 \ \\ \ \\ \ \\ 2x^2 -10x +12 = 0 \ \\ \ \\ a = 2; b = -10; c = 12 \ \\ D = b^2 - 4ac = 10^2 - 4\cdot 2 \cdot 12 = 4 \ \\ D>0 \ \\ \ \\ x_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 10 \pm \sqrt{ 4 } }{ 4 } \ \\ x_{1,2} = \dfrac{ 10 \pm 2 }{ 4 } \ \\ x_{1,2} = 2.5 \pm 0.5 \ \\ x_{1} = 3 \ \\ x_{2} = 2 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 2 (x -3) (x -2) = 0 \ \\ x_{ 1 } = 3

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x2=2x_{ 2 } = 2

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