What are the solutions to the equation x^2+4b=0,b<0 ?

The solutions to the equation x^2+4b=0,b<0 are x=2sqrt(-b),x=-2sqrt(-b)

Find the zeros of x^2+4b=0,b<0

The zeros of x^2+4b=0,b<0 are x=2sqrt(-b),x=-2sqrt(-b)

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The solutions to the equation x^2+4b=0,b<0 are x=2sqrt(-b),x=-2sqrt(-b)

The zeros of x^2+4b=0,b<0 are x=2sqrt(-b),x=-2sqrt(-b)

x^{2} + 4 b = 0, b < 0

x = 2 - b, x = - 2 - b

Solution steps

x^{2} + 4 b = 0, b < 0

Move

4 b

to the right side

x^{2} = - 4 b

For

x^{2} = f (a)

the solutions are

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

x = - 4 b, x = - - 4 b

Simplify

- 4 b : 2 - b

Simplify

- - 4 b : - 2 - b

x = 2 - b, x = - 2 - b

(2 x - 1) (2 + x) = (x - 2) (x + 1)

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

(2 x + 3) (3 x - 7) = 0

3 (x + 4)^{2} = 8

- 4 + 16 x - 15 x^{2} = 0

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