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

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

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

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

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

x = 10, x = 2

Solution steps

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

Expand

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

\frac{\frac{2}{3}}{4}

x - 7 = - 10

- 3 \leq 5 + 2 x < 5

(p - 7)^{2} - 10 = 0

4 \times \frac{5}{8}

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