What are the solutions to the equation (x-1)^2+(0-4)^2=36 ?

The solutions to the equation (x-1)^2+(0-4)^2=36 are x=2sqrt(5)+1,x=-2sqrt(5)+1

Find the zeros of (x-1)^2+(0-4)^2=36

The zeros of (x-1)^2+(0-4)^2=36 are x=2sqrt(5)+1,x=-2sqrt(5)+1

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(x - 1)^{2} + (0 - 4)^{2} = 36

x = 25 + 1, x = - 25 + 1

Decimal Notation

x = 5.47213 \dots, x = - 3.47213 \dots

Solution steps

(x - 1)^{2} + (0 - 4)^{2} = 36

Move

(0 - 4)^{2}

to the right side

(x - 1)^{2} = 20

For

(g (x))^{2} = f (a)

the solutions are

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

Solve

x - 1 = 20 : x = 25 + 1

Solve

x - 1 = - 20 : x = - 25 + 1

The solutions to the quadratic equation are:

x = 25 + 1, x = - 25 + 1

4 ∣ x - 2 ∣ = 32

41 k = 3 (- 41 k + 3)

\frac{x}{∣ x - 2 ∣} > 2

(2^{\frac{1}{2}})^{2}

5 x + 8 = 0

What are the solutions to the equation (x-1)^2+(0-4)^2=36 ?
The solutions to the equation (x-1)^2+(0-4)^2=36 are x=2sqrt(5)+1,x=-2sqrt(5)+1
Find the zeros of (x-1)^2+(0-4)^2=36
The zeros of (x-1)^2+(0-4)^2=36 are x=2sqrt(5)+1,x=-2sqrt(5)+1