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

The solutions to the equation 2^2+4^2=x^2 are x=2sqrt(5),x=-2sqrt(5)

Find the zeros of 2^2+4^2=x^2

The zeros of 2^2+4^2=x^2 are x=2sqrt(5),x=-2sqrt(5)

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

x = 25, x = - 25

Decimal Notation

x = 4.47213 \dots, x = - 4.47213 \dots

Solution steps

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

Switch sides

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

Simplify

2^{2} + 4^{2} : 20

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = 20, x = - 20

20 = 25

- 20 = - 25

x = 25, x = - 25

8 x^{2} = 72

x^{2} - 6 x + 9 = 64

x^{2} + 7 x + 15 = 0

(x + 7)^{2} = 9

3 x^{2} + 4 x + 7 = - 1

What are the solutions to the equation 2^2+4^2=x^2 ?
The solutions to the equation 2^2+4^2=x^2 are x=2sqrt(5),x=-2sqrt(5)
Find the zeros of 2^2+4^2=x^2
The zeros of 2^2+4^2=x^2 are x=2sqrt(5),x=-2sqrt(5)