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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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

x^{2} - 7 x + \frac{5}{4} = 0

3 q^{2} - 105 q + 750 = 0

z^{2} - 7 z + 13 = 0

m x^{2} - 5 x + 2 = n x - 1

(2 x - 1) (x + 3) = 4

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)