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

Answer 1

Eva Abramson

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

Adding #4# to both ends we find:

#(x-1)^2 = 4#

So #x-1 = +-sqrt(4) = +-2#

Adding #1# to both sides, we get:

#x = 1+-2#

That is #x = -1# or #x = 3#

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Answer 2

Eva Abramson

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

Move the constant term to the other side of the equation: x^2 - 2x = 3.
To complete the square, add the square of half the coefficient of x to both sides of the equation: x^2 - 2x + (2/2)^2 = 3 + (2/2)^2.
Simplify: x^2 - 2x + 1 = 3 + 1.
Rewrite the left side as a squared binomial: (x - 1)^2 = 4.
Take the square root of both sides: x - 1 = ±√4.
Solve for x: x - 1 = ±2.
Add 1 to both sides: x = 1 ± 2.

Therefore, the solutions are x = 1 + 2 and x = 1 - 2, which simplifies to x = 3 and x = -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.