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

The solutions to the equation 5^2+4^2=x^2 are x=sqrt(41),x=-sqrt(41)

Find the zeros of 5^2+4^2=x^2

The zeros of 5^2+4^2=x^2 are x=sqrt(41),x=-sqrt(41)

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5^{2} + 4^{2} = x^{2}

x = 41, x = - 41

Decimal Notation

x = 6.40312 \dots, x = - 6.40312 \dots

Solution steps

5^{2} + 4^{2} = x^{2}

Switch sides

x^{2} = 5^{2} + 4^{2}

Simplify

5^{2} + 4^{2} : 41

For

x^{2} = f (a)

the solutions are

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

x = 41, x = - 41

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

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

a, f (a + 1) = (a + 1)^{2} + 3 (a + 1) - 4

x^{2} - 22 x + 57 = 0

4 x^{2} - 40 = 0

What are the solutions to the equation 5^2+4^2=x^2 ?
The solutions to the equation 5^2+4^2=x^2 are x=sqrt(41),x=-sqrt(41)
Find the zeros of 5^2+4^2=x^2
The zeros of 5^2+4^2=x^2 are x=sqrt(41),x=-sqrt(41)