How do you complete the square to solve #-x^2-2x+3= 0#?

Answer 1

#(x+1)^2-4=0#
Solutions: #x=1 or x=-3#

If you have to solve the equation you can multiply all by -1: #−x^2−2x+3=0 -> x^2+2x-3=0# It's always more simple to work with positive numbers =) # x^2+2x-3=0# is similar to a perfect square trinomial that is #x^2+2x+1=(x+1)^2#. So we can write our equation adding (or subtracting) what is missing to obtain a perfect square: # x^2+2x-3=(x+1)^2-4=0# If you want to solve the equation you have to obtain 4 in the square #(x+1)^2=4#, so #x+1=+-2# and the solutions are: #x=1 or x=-3#
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Answer 2

#x=1, -3#

A perfect square trinomial is in the form #(a+b)^2=a^2+2ab+b^2#.
#-x^2-2x+3=0#
Multiply the equation times #-1#.
#x^2+2x-3=0#
The left side of the equation can be made into a perfect square trinomial by completing the square. This will enable the equation to be solved for #x#.
Add #3# to both sides.
#x^2+2x=3#
Divide the coefficient of the #x# term by #2#, then square the result. Add it to both sides of the equation.
#(2/2)^2=1#
#x^2+2x+1=4#

The left side of the equation now has a perfect square trinomial.

#(a+b)^2=a^2+2ab+b^2#
#a=x# #b=1#
#(x+1)^2=4#

Take each side's square root.

#x+1=+-sqrt4#
#x+1=+-2#
If #x+1=2#, #x=1#.
If #x+1=-2#, #x=-3#
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Answer 3

To solve the quadratic equation (-x^2 - 2x + 3 = 0) by completing the square, follow these steps:

  1. Move the constant term to the other side of the equation:

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

  1. Make sure the coefficient of the (x^2) term is 1. If it's not, divide all terms by the coefficient of (x^2). In this case, the coefficient is already -1, so no change is needed.

  2. Rewrite the equation with space for adding the square of half the coefficient of (x). The coefficient of (x) is -2, so half of it is -1. Square this value to get 1, and add it to both sides of the equation:

(-x^2 - 2x + 1 = -3 + 1)

  1. Factor the left side of the equation:

(-(x + 1)^2 = -2)

  1. Divide both sides of the equation by -1 to isolate the squared term:

((x + 1)^2 = 2)

  1. Take the square root of both sides:

[x + 1 = \pm \sqrt{2}]

  1. Solve for (x):

[x = -1 \pm \sqrt{2}]

So, the solutions to the equation (-x^2 - 2x + 3 = 0) are (x = -1 + \sqrt{2}) and (x = -1 - \sqrt{2}).

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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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