What are the solutions to the equation (x-2)^2+(x-4)^2=x^2 ?

The solutions to the equation (x-2)^2+(x-4)^2=x^2 are x=10,x=2

Find the zeros of (x-2)^2+(x-4)^2=x^2

The zeros of (x-2)^2+(x-4)^2=x^2 are x=10,x=2

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(x - 2)^{2} + (x - 4)^{2} = x^{2}

x = 10, x = 2

Solution steps

(x - 2)^{2} + (x - 4)^{2} = x^{2}

Expand

(x - 2)^{2} + (x - 4)^{2} : 2 x^{2} - 12 x + 20

2 x^{2} - 12 x + 20 = x^{2}

Move

x^{2}

to the left side

x^{2} - 12 x + 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 20 = 8

x_{1, 2} = \frac{- ( - 12 ) \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 8}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 8}{2 \cdot 1}

x = \frac{- ( - 12 ) + 8}{2 \cdot 1} : 10

x = \frac{- ( - 12 ) - 8}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 10, x = 2

t^{2} + 10 t + 9 = 0

11 x^{2} - 153 x - 180 = 0

a^{2} + a + 1 = 0

\frac{c ^{2}}{8} - 3 = 2

x (2 x - 3) = 20

What are the solutions to the equation (x-2)^2+(x-4)^2=x^2 ?
The solutions to the equation (x-2)^2+(x-4)^2=x^2 are x=10,x=2
Find the zeros of (x-2)^2+(x-4)^2=x^2
The zeros of (x-2)^2+(x-4)^2=x^2 are x=10,x=2