What are the solutions to the equation (32-x)^2+(16-x)^2=x^2 ?

The solutions to the equation (32-x)^2+(16-x)^2=x^2 are x=80,x=16

Find the zeros of (32-x)^2+(16-x)^2=x^2

The zeros of (32-x)^2+(16-x)^2=x^2 are x=80,x=16

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(32 - x)^{2} + (16 - x)^{2} = x^{2}

x = 80, x = 16

Solution steps

(32 - x)^{2} + (16 - x)^{2} = x^{2}

Expand

(32 - x)^{2} + (16 - x)^{2} : 2 x^{2} - 96 x + 1280

2 x^{2} - 96 x + 1280 = x^{2}

Move

x^{2}

to the left side

x^{2} - 96 x + 1280 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 96 ) \pm ( - 96 ) ^{2} - 4 \cdot 1 \cdot 1280}{2 \cdot 1}

(- 96)^{2} - 4 \cdot 1 \cdot 1280 = 64

x_{1, 2} = \frac{- ( - 96 ) \pm 64}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 96 ) + 64}{2 \cdot 1}, x_{2} = \frac{- ( - 96 ) - 64}{2 \cdot 1}

x = \frac{- ( - 96 ) + 64}{2 \cdot 1} : 80

x = \frac{- ( - 96 ) - 64}{2 \cdot 1} : 16

The solutions to the quadratic equation are:

x = 80, x = 16

5 x^{2} + 40 x = - 75

x^{2} - x + \frac{1}{8} = 0

x, 0 = x (1 - x) + m - h x

6 x^{2} - 5 x + 4 = 4 (3 x^{2} - 2 x + 1)

t, h = - 16 t^{2} + v^{0} t + h^{0}

What are the solutions to the equation (32-x)^2+(16-x)^2=x^2 ?
The solutions to the equation (32-x)^2+(16-x)^2=x^2 are x=80,x=16
Find the zeros of (32-x)^2+(16-x)^2=x^2
The zeros of (32-x)^2+(16-x)^2=x^2 are x=80,x=16