How to make 2x^2-4x=-3 a perfect square

The original 2x^2 - 4x = - 3 becomes 2[ (x - 1) ^2 - 1 = - 3.

Step-by-step explanation:

Start with 2x^2-4x=-3. Rewrite this as 2x^2-4x = - 3, and then as

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

Next, "complete the square" of x^2 - 2x: Take half of the coefficient (-2) of x (which comes out to - 1). Square this result (obtaining (-1) ^2, or just 1. Add this 1 to x^2 - 2x, and then subtract the same quantity (1) : x^2 - 2x + 1 - 1.

Now rewrite the perfect square x^2 - 2x + 1 as (x - 1) ^2, and then subtract that - 1 (above):

(x - 1) ^2 - 1.

Substituting this for x^2 - 2x in 2 (x^2 - 2x) = - 3, we get:

2 [ (x - 1) ^2 - 1 ] = - 3, or

2 (x - 1) ^2 - 2 = - 3, or

2 (x - 1) ^2 = - 1

We weren't asked to solve this equation, but may as well do so:

Divide both sides by 2, obtaining (x - 1) ^2 = - 1/2

Normally, we'd find the square root of both sides, but here we cannot find the square root of the negative number - 1/2, unless complex roots are acceptable.