Question

how do you solve 4*x^2 + 49 = 0

Answer 1

\[4x^{2} + 49 =0\]

Answer 2

I'm sure there are other ways, but you can do:

4x^2 + 49 = 0
4x^2 = -49
x^2 = -49 / 4
x = sqrt ( -49 / 4)

which results in a complex solution. There is probably a nicer way to do it.

Answer 3

4x^2+49=0
4x^2=-49
4x^2=49i^2
x^2=(49/4)*i^2
x=(7/2)i
complex number for i

Answer 4

after i subtract 49 to both sides do i devide by 4 first? or square root?

Answer 5

Divide by 4 first

Answer 6

im left with x\[x ^{2} = 49\div4\]

Answer 7

you left the minus sign

Answer 8

-49

Answer 9

correct (except for the minus sign), now take the square root of both sides, which solves for x
so you get

x =sqrt ( -49/4 )

Answer 10

should be x^2= -49/4 instead

Answer 11

after that how would i square root?

Answer 12

Notice that 49 is 7^2 and 4 is 2^2, so

sqrt ( -49 / 4) = 7/2, however you are taking the square root of a negative number, so you need to multiply by a complex factor i.

So sqrt ( -49 / 4) = 7/2i, to be precise

Answer 13

i need imaginary numbers if any

Answer 14

i^2 = -1

Answer 15

the answer would be \[\pm7i/2?\]

Answer 16

it does not need to be plus or minus, otherwise the answer is correct

Answer 17

the i takes care of the plus or minus

Answer 18

but when u squareroot an equation isn't there always plus or minus?

Answer 19

the answer is just imaginary.... not a real number.

Answer 20

If you are going to include i, you do not need to have a plus or minus attached to the square root. To do so would be redundant.

If you take the square root of -1, you get i.
So the square root of -49, is 7i.

It actually doesn't make sense to say sqrt (-1) = +- i

Answer 21

:)