What are the solutions to the equation x^2+x^2=(x+6)^2 ?

The solutions to the equation x^2+x^2=(x+6)^2 are x=-6(sqrt(2)-1),x=6(1+sqrt(2))

Find the zeros of x^2+x^2=(x+6)^2

The zeros of x^2+x^2=(x+6)^2 are x=-6(sqrt(2)-1),x=6(1+sqrt(2))

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x^{2} + x^{2} = (x + 6)^{2}

x = - 6 (2 - 1), x = 6 (1 + 2)

Solution steps

x^{2} + x^{2} = (x + 6)^{2}

Expand

(x + 6)^{2} : x^{2} + 12 x + 36

2 x^{2} = x^{2} + 12 x + 36

Switch sides

x^{2} + 12 x + 36 = 2 x^{2}

Move

2 x^{2}

to the left side

- x^{2} + 12 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 1 ) \cdot 36}{2 ( - 1 )}

1 2^{2} - 4 (- 1) \cdot 36 = 122

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

Separate the solutions

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

x = \frac{- 12 + 12 2}{2 ( - 1 )} : - 6 (2 - 1)

x = \frac{- 12 - 12 2}{2 ( - 1 )} : 6 (1 + 2)

The solutions to the quadratic equation are:

x = - 6 (2 - 1), x = 6 (1 + 2)

2 (w + 1) + 4 w = 3 (2 w - 1) + 8

(t^{4})^{2}

5 + \frac{28}{x} \geq 0

x \leq 4 and x > - 2

- \frac{1}{4} - 1

What are the solutions to the equation x^2+x^2=(x+6)^2 ?
The solutions to the equation x^2+x^2=(x+6)^2 are x=-6(sqrt(2)-1),x=6(1+sqrt(2))
Find the zeros of x^2+x^2=(x+6)^2
The zeros of x^2+x^2=(x+6)^2 are x=-6(sqrt(2)-1),x=6(1+sqrt(2))