How do you solve #2y^2+6=y^2#?

Answer 1
#y=sqrt(6)i=isqrt6#
Problem: Solve #2y^2+6=y^2#.
Subtract #6# from both sides of the equation.
#2y^=y^2-6#
Subtract #y^2# from both sides.
#y^2=-6#

Take the square root of both sides.

#y=sqrt(-6)# =
#y=sqrt(6)i=isqrt6#
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Answer 2

To solve the equation 2y^2 + 6 = y^2:

  1. Subtract y^2 from both sides: 2y^2 + 6 - y^2 = y^2 - y^2 y^2 + 6 = 0

  2. Subtract 6 from both sides: y^2 + 6 - 6 = 0 - 6 y^2 = -6

  3. Take the square root of both sides: √(y^2) = √(-6) y = ±√(-6)

Since the square root of a negative number is imaginary, there are no real solutions to this equation. The solutions are complex numbers: y = ±√(6)i.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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