What are the solutions to the equation x^2+(4+x)^2=(8+x)^2 ?

The solutions to the equation x^2+(4+x)^2=(8+x)^2 are x=12,x=-4

Find the zeros of x^2+(4+x)^2=(8+x)^2

The zeros of x^2+(4+x)^2=(8+x)^2 are x=12,x=-4

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x^{2} + (4 + x)^{2} = (8 + x)^{2}

x = 12, x = - 4

Solution steps

x^{2} + (4 + x)^{2} = (8 + x)^{2}

Expand

x^{2} + (4 + x)^{2} : 2 x^{2} + 8 x + 16

Expand

(8 + x)^{2} : 64 + 16 x + x^{2}

2 x^{2} + 8 x + 16 = 64 + 16 x + x^{2}

Move

x^{2}

to the left side

x^{2} + 8 x + 16 = 64 + 16 x

Move

16 x

to the left side

x^{2} - 8 x + 16 = 64

Move

64

to the left side

x^{2} - 8 x - 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot ( - 48 )}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot (- 48) = 16

x_{1, 2} = \frac{- ( - 8 ) \pm 16}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 16}{2 \cdot 1}, x_{2} = \frac{- ( - 8 ) - 16}{2 \cdot 1}

x = \frac{- ( - 8 ) + 16}{2 \cdot 1} : 12

x = \frac{- ( - 8 ) - 16}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 12, x = - 4

2 x^{2} - 16 x + 24 = 0

2 x^{2} + 20 x - 22 = 0

2 x^{2} - 28 x + 102

x, x^{2} + 2 x - 24 = 0

15 x = x^{2} - 16

What are the solutions to the equation x^2+(4+x)^2=(8+x)^2 ?
The solutions to the equation x^2+(4+x)^2=(8+x)^2 are x=12,x=-4
Find the zeros of x^2+(4+x)^2=(8+x)^2
The zeros of x^2+(4+x)^2=(8+x)^2 are x=12,x=-4