For example:

25y^2+16=0

February 2nd, 2011 @2:48 pm

with complex numbers

February 2nd, 2011 @3:30 pm

that’s not a quadratic equation.

this is easy to solve

subtract 16 from both sides of equation

25y^2 = -16

divide both sides by 25

y^2 = -16/25

take the square root of both sides

y = sqrt(-16/25)

y = -4/5

y = -0.8

February 2nd, 2011 @3:49 pm

25y^2 + 16 = 0

25y^2 = -16

y^2 = -16/25

sqrt(y^2) = sqrt(-16/25)

y = sqrt(-16/25)

sqrt = square root

February 2nd, 2011 @4:21 pm

one way for a problem like this (and it’s an equation, not a formula)

use the square root property

25y^2 + 16 = 0

25y^2 = -16

y^2 = -16 / 25 (no real solutions, so if you haven’t learned imaginary roots, stop here)

y = +/- sqrt(-16 / 25) = +/- 4i / 5 (where i^2 = -1)

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or, if you really must use the quadratic formula

given ay^2 + by + c = 0, then

y = [-b +/- sqrt(b^2 - 4ac)] / (2a)

here, a = 25, b = 0, and c = 16

y = [-0 +/- sqrt(0 - 4(25)(16))] / (2*25)

y = +/- sqrt(-1600) / 50 = +/- i 40 / 50

y = +/- 4i /5 (looks familiar)

February 2nd, 2011 @4:45 pm

See: http://www.purplemath.com/modules/quadform.htm

Write it as 25y^2 + 0y + 16 = 0 and apply the quadratic formula for ax^2 + by + c

February 2nd, 2011 @5:04 pm

25y² + 16 = 0

25y² = – 16

y² = – 16/25

y = ± 16i/5

Answer: y = ± 16i/5—imaginary solutions only.

February 2nd, 2011 @5:28 pm

there is no solution to this problem because there is no value to which y can equall 0 . Confifmed by the quadratic formula -b(+-)(b^2-4ac)^1/2 the values produced are not real therefore no solution

2a

February 2nd, 2011 @6:14 pm

25 y ² = – 16

y ² = – (16 / 25 )

y ² = i ² ( 16 / 25 )

y = ± (4/5) i

x= -b(plus an minus sign) the sqareroot of [(b^2-4ac)/2a]