How do you solve #x^2-2x+6=0# by completing the square?

Answer 1
Write it as: #x^2-2x=-6# Add and subtract #1#; #x^2-2x+1-1=-6# #x^2-2x+1=-6+1# #(x-1)^2=-5# So: #x-1=+-sqrt(-5)# #x=1+-isqrt(5)# These are two Complex Solutions where #i=sqrt(-1)#
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Answer 2

To solve the equation x^2 - 2x + 6 = 0 by completing the square, follow these steps:

  1. Move the constant term to the other side of the equation: x^2 - 2x = -6

  2. Take half of the coefficient of x, square it, and add it to both sides of the equation: x^2 - 2x + (-2/2)^2 = -6 + (-2/2)^2 x^2 - 2x + 1 = -6 + 1 x^2 - 2x + 1 = -5

  3. Rewrite the left side of the equation as a perfect square trinomial and simplify the right side: (x - 1)^2 = -5

  4. Take the square root of both sides: x - 1 = ±√(-5)

  5. Add 1 to both sides: x = 1 ± √(-5)

  6. Since the square root of a negative number is imaginary, the solution is complex: x = 1 ± i√5

So, the solutions to the equation x^2 - 2x + 6 = 0 are x = 1 + i√5 and x = 1 - i√5.

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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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