How do you solve the equation #x^2+a=4#?

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Answer 1 · 2017-07-26T07:25:52

Arrange your equation

Arrange your equation

#x^2 = 4-a#

#x = sqrt(4-a)#

a must be less than or equal to 4.

#x=sqrt(4-a)#

If a is 1, for instance, you can get:

#x=sqrt3#