How do you solve by completing the square for #5-4x-x^2#?

Answer 1

I assume you meant to solve #5-4x-x^2 = 0#

This is equivalent to #color(white)("XXXXX")##x^2+4x = 5# Completing the square #color(white)("XXXXX")##x^2+4x+4 = 5+4# #color(white)("XXXXX")##(x+2)^2 = 9# Take the square root #color(white)("XXXXX")##x+2 = +-3# Solve for x #color(white)("XXXXX")##x=-2+-3#

#x=-5# or #x=+1#

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Answer 2

Eva Adamson

To solve the quadratic equation (5 - 4x - x^2) by completing the square:

Rewrite the equation in the form (x^2 + bx + c): (x^2 - 4x + 5).
Identify the coefficient of (x), which is (b = -4).
Compute ((b/2)^2): ((-4/2)^2 = (-2)^2 = 4).
Add and subtract the result from step 3 inside the parentheses: (x^2 - 4x + 4 - 4 + 5).
Rewrite the expression as a perfect square trinomial and a constant term: ((x - 2)^2 - 4 + 5).
Simplify: ((x - 2)^2 + 1).

So, the equation (5 - 4x - x^2) when solved by completing the square is ((x - 2)^2 + 1).

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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.