Distance problem 2

A=(x,2x)
B=(2x,1)
Distance AB=√2, find value of x

Correct result:

x1 =  1
x2 =  -0.2

Solution:

(x−2x)2+(2x−1)2=2  (x−2x)2+(2x−1)2=2  5x2−4x−1=0  a=5;b=−4;c=−1 D=b2−4ac=42−4⋅5⋅(−1)=36 D>0  x1,2=−b±D2a=4±3610 x1,2=4±610 x1,2=0.4±0.6 x1=1 x2=−0.2   Factored form of the equation:  5(x−1)(x+0.2)=0  x1=1(x-2x)^2 + (2x-1)^2=2 \ \\ \ \\ (x-2x)^2 + (2x-1)^2=2 \ \\ \ \\ 5x^2 -4x -1=0 \ \\ \ \\ a=5; b=-4; c=-1 \ \\ D=b^2 - 4ac=4^2 - 4\cdot 5 \cdot (-1)=36 \ \\ D>0 \ \\ \ \\ x_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 4 \pm \sqrt{ 36 } }{ 10 } \ \\ x_{1,2}=\dfrac{ 4 \pm 6 }{ 10 } \ \\ x_{1,2}=0.4 \pm 0.6 \ \\ x_{1}=1 \ \\ x_{2}=-0.2 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 5 (x -1) (x +0.2)=0 \ \\ \ \\ x_{1}=1

Checkout calculation with our calculator of quadratic equations.

x2=(−0.2)=−15=−0.2x_{2}=(-0.2)=- \dfrac{ 1 }{ 5 }=-0.2

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