What are the solutions to the equation (38-x)^2+(19-x)^2=x^2 ?

The solutions to the equation (38-x)^2+(19-x)^2=x^2 are x=95,x=19

Find the zeros of (38-x)^2+(19-x)^2=x^2

The zeros of (38-x)^2+(19-x)^2=x^2 are x=95,x=19

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(38 - x)^{2} + (19 - x)^{2} = x^{2}

x = 95, x = 19

Solution steps

(38 - x)^{2} + (19 - x)^{2} = x^{2}

Expand

(38 - x)^{2} + (19 - x)^{2} : 2 x^{2} - 114 x + 1805

2 x^{2} - 114 x + 1805 = x^{2}

Move

x^{2}

to the left side

x^{2} - 114 x + 1805 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 114 ) \pm ( - 114 ) ^{2} - 4 \cdot 1 \cdot 1805}{2 \cdot 1}

(- 114)^{2} - 4 \cdot 1 \cdot 1805 = 76

x_{1, 2} = \frac{- ( - 114 ) \pm 76}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 114 ) + 76}{2 \cdot 1}, x_{2} = \frac{- ( - 114 ) - 76}{2 \cdot 1}

x = \frac{- ( - 114 ) + 76}{2 \cdot 1} : 95

x = \frac{- ( - 114 ) - 76}{2 \cdot 1} : 19

The solutions to the quadratic equation are:

x = 95, x = 19

x^{2} + 10 x + 20 = 0

7 x^{2} - x + 8 = 0

(x + 1)^{2} = 8

5 x^{2} = 27 x + 18

x, x^{2} - y^{2} - 5 x y - 1 = 0

What are the solutions to the equation (38-x)^2+(19-x)^2=x^2 ?
The solutions to the equation (38-x)^2+(19-x)^2=x^2 are x=95,x=19
Find the zeros of (38-x)^2+(19-x)^2=x^2
The zeros of (38-x)^2+(19-x)^2=x^2 are x=95,x=19