What is the answer to 10=sqrt((0-k)^2+(2k-0)^2) ?

The answer to 10=sqrt((0-k)^2+(2k-0)^2) is k=sqrt(20),k=-sqrt(20)

Solve for k: 10=sqrt((0-k)^2+(2k-0)^2)

The solution is k=sqrt(20),k=-sqrt(20)

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10 = (0 - k)^{2} + (2 k - 0)^{2}

k = 20, k = - 20

Decimal Notation

k = 4.47213 \dots, k = - 4.47213 \dots

Solution steps

10 = (0 - k)^{2} + (2 k - 0)^{2}

Square both sides:

100 = 5 k^{2}

100 = 5 k^{2}

Solve

100 = 5 k^{2} : k = 20, k = - 20

k = 20, k = - 20

Verify Solutions:

k = 20

True

, k = - 20

True

The solutions are

k = 20, k = - 20

x + 3 + x - 2 = 5

a = x^{2} - 1

- 1 = (- 2)^{2} + (- 2) + c_{1}

x - 1 = 3 x - 13

What is the answer to 10=sqrt((0-k)^2+(2k-0)^2) ?
The answer to 10=sqrt((0-k)^2+(2k-0)^2) is k=sqrt(20),k=-sqrt(20)
Solve for k: 10=sqrt((0-k)^2+(2k-0)^2)
The solution is k=sqrt(20),k=-sqrt(20)