Distance problem

A=(x, x)
B=(1,4)
Distance AB=√5, find x;

Result

x1 =  3
x2 =  2

Solution:

(x−1)2+(x−4)2=5   2x2−10x+12=0  a=2;b=−10;c=12 D=b2−4ac=102−4⋅2⋅12=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(x−3)(x−2)=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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