Performing Arithmetic with Complex Numbers

To solve the given equation, let's first isolate

x^{2} .

2 x^{2} + 50 = 0

LHS - 50 = RHS - 50

2 x^{2} = - 50

LHS / 2 = RHS / 2

x^{2} = - 25

We cannot find any real square roots of a negative number, so we will need to use the fact that

- a = i a

to solve for

x .

x^{2} = - 25

LHS = RHS

x = \pm - 25

- a = i a

x = \pm i 25

CalcRootCalculate root

x = \pm i \cdot 5

x = \pm 5 i

Performing Arithmetic with Complex Numbers 1.5 - Solution