how to solve this qudratic equation plz?

Question

4x^2 + 12=0
help me plz

Answer 1

Your equation has no real roots. The two roots are imaginary:

x² + 3 = 0
x² = -3
x = ± i√3

Answer 2

You can solve it in this manner

4x^2 + 12 = 0
4x^2 = -12
x^2 = -3
x = sqrt(-3) = sqrt(3) i

Remember in maths, the sqrt value can be positive or negative. Therefore the solutions are x = sqrt(3) i
and x = -sqrt(3) i

Btw. in reality, a negative number does not have a square root. Therefore, mathematicians have introduced the concept of i, where i^2 = -1. So, to further explain how i got 'i' above is shown here

sqrt(-3) = sqrt( -1 * 3) = sqrt(-1) * sqrt(3) = i * sqrt(3)


Hope this helps

Answer 3

4x^2 + 12 = 0
4x^2 = -12
x^2 = -12/4 -------- Divide by 4 throughout

x^2 = -3
x = Sqrt(-3)

x has no real solution becos there is no real solution to square root of negative number.

Answer 4

Subtract 12 from both sides
4x^2=-12
Divide by 4
x^2=-3
Square root both sides
sqrt(x^2)=sqrt(-3)
If you're in algebra 1 at this point you'd say No Real Solution.
For algebra 2 continue on,
x=sqrt(3)i, x=-sqrt(3)i

Answer 5

Factor it into 4(x^2 + 3) = 0, then x^2 has to be -3, and you cannot get negative numbers using an even exponent, so therefore the equation has no solution, unless you are dealing with imaginary numbers, then the answer is + or - i root3

Answer 6

Alrighty. right here we circulate. 3(x-4) = 2x - (6+x) 3x - 12 = 2x - 6 - x See how the destructive in front of the parenthesee modifications the sign of the values interior it?? this is a math rule. A destructive (minus sign) in front of a few thing interior parentheses is comparable to multiplying each and every thing in there by -a million, undergo in suggestions. So we've 3x - 12 = x - 6 including 6 to the two area, we get 3x - 6 = x Subtract x from the two area 2x - 6 = 0 upload 6 to the two area 2x = 6 x = 3

Answer 7

Solving for x^2 gives x^2=-3 which has x=isqrt(3) and x=-isqrt(3) as solutions.

Answer 8

4x^2 = -12
x^2 = -12/4 = -3
x = + or - sqrt(-3) = +sqrt(3)i or -sqrt(3)i
where i = sqrt(-1)