How do you solve #-2m ^ { 2} - 6= 0#?

Question

Kennedy Adams · Accepted Answer

There are no real solutions.

#-2m^2 -6=0# Although this is a quadratic # (m^2)# we can use a different method to solve it because there is no #m# term. Write it as #" "2m^2 = -6" "larr# solve for #m. div 2# #m^2 = -3" "larr# find the square root #m = +-sqrt-3# We now have a problem because a negative value cannot have its square root found. As a result, this equation has no real solutions.

Isaiah Anthony · Answer

undefined To solve the equation -2m^2 - 6 = 0, we can start by adding 6 to both sides of the equation to isolate the term with the variable. This gives us -2m^2 = 6.

Next, we divide both sides of the equation by -2 to solve for m^2. This yields m^2 = -3.

To find the value of m, we take the square root of both sides of the equation. However, it's important to note that when dealing with square roots, we need to consider both the positive and negative square roots.

Taking the square root of m^2 = -3 gives us m = ±√(-3).

Since the square root of a negative number is not a real number, this equation has no real solutions.

Genesis Allen · Answer

undefined To solve the equation $-2m^2 - 6 = 0$, follow these steps:

1. Add 6 to both sides of the equation: $-2m^2 = 6$.
2. Divide both sides by $-2$: $m^2 = -3$.
3. Take the square root of both sides. Remember to consider both positive and negative roots: $m = \pm \sqrt{-3}$.
4. Since the square root of a negative number is not a real number, the solutions are imaginary: $m = \pm \sqrt{3}i$.