Distance problem

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

Correct result:

x₁ =  3
x₂ =  2

Solution:

$$\begin{gathered} (x-1{)}^2+(x-4{)}^2=5 \\ (x-1{)}^2+(x-4{)}^2=5 \\ 2x^2-10x+12=0 \\ a=2;b=-10;c=12 \\ D=b^2-4ac=10^2-4\cdot 2\cdot 12=4 \\ D>0 \\ {x}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{10\pm \sqrt{4}}{4}} \\ {x}_{1,2}={\displaystyle \frac{10\pm 2}{4}} \\ {x}_{1,2}=2.5\pm 0.5 \\ {x}_1=3 \\ {x}_2=2 \\ \text{ Factored form of the equation: } \\ 2(x-3)(x-2)=0 \\ {x}_1=3 \end{gathered}$$

Our quadratic equation calculator calculates it.

$x_{2}=2$

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