What are the solutions to the equation x^2+(2x)^2=5 ?

The solutions to the equation x^2+(2x)^2=5 are x=1,x=-1

Find the zeros of x^2+(2x)^2=5

The zeros of x^2+(2x)^2=5 are x=1,x=-1

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x^{2} + (2 x)^{2} = 5

x = 1, x = - 1

Solution steps

x^{2} + (2 x)^{2} = 5

Expand

x^{2} + (2 x)^{2} : 5 x^{2}

5 x^{2} = 5

Move

5

to the left side

5 x^{2} - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

0^{2} - 4 \cdot 5 (- 5) = 10

x_{1, 2} = \frac{- 0 \pm 10}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 0 + 10}{2 \cdot 5}, x_{2} = \frac{- 0 - 10}{2 \cdot 5}

x = \frac{- 0 + 10}{2 \cdot 5} : 1

x = \frac{- 0 - 10}{2 \cdot 5} : - 1

The solutions to the quadratic equation are:

x = 1, x = - 1

5 w^{2} - 8 w - 7 = 0

2 mn x^{2} + n x - 2 m x - 1 = 0

x^{2} + (\frac{4}{5})^{2} = 1

0 = y^{2} - 4 y

(\frac{3}{2} x - \frac{1}{2})^{2} - 0.8 (1 - x) = 0

What are the solutions to the equation x^2+(2x)^2=5 ?
The solutions to the equation x^2+(2x)^2=5 are x=1,x=-1
Find the zeros of x^2+(2x)^2=5
The zeros of x^2+(2x)^2=5 are x=1,x=-1