What are the solutions to the equation x^2+(2x)^2=4sqrt(5) ?

The solutions to the equation x^2+(2x)^2=4sqrt(5) are x=-2/(\sqrt[4]{5)},x= 2/(\sqrt[4]{5)}

Find the zeros of x^2+(2x)^2=4sqrt(5)

The zeros of x^2+(2x)^2=4sqrt(5) are x=-2/(\sqrt[4]{5)},x= 2/(\sqrt[4]{5)}

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x^{2} + (2 x)^{2} = 45

Decimal Notation

x = - 1.33748 \dots, x = 1.33748 \dots

Solution steps

x^{2} + (2 x)^{2} = 45

Expand

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

5 x^{2} = 45

Move

45

to the left side

5 x^{2} - 45 = 0

Factor

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

Solve

The solutions to the quadratic equation are:

B^{2} - 81 = 0

x^{2} - 3 x + 2

x^{2} + 5 x + c

x^{2} + 3 x = - 4 x

s^{2} + 8 s + 16

What are the solutions to the equation x^2+(2x)^2=4sqrt(5) ?
The solutions to the equation x^2+(2x)^2=4sqrt(5) are x=-2/(\sqrt[4]{5)},x= 2/(\sqrt[4]{5)}
Find the zeros of x^2+(2x)^2=4sqrt(5)
The zeros of x^2+(2x)^2=4sqrt(5) are x=-2/(\sqrt[4]{5)},x= 2/(\sqrt[4]{5)}