Question

solve each by completing the square.

x^2+2x-3=0

Answer 1

Solve for x over the real numbers:
\[x^2+2 x-3 = 0 \]Solve the quadratic equation by completing the square.
Add 3 to both sides:
\[x^2+2 x = 3 \]Take one half of the coefficient of x and square it, then add it to both sides.
Add 1 to both sides:
\[x^2+2 x+1 = 4 \]Factor the left hand side.
Write the left hand side as a square:
\[(x+1)^2 = 4 \]Eliminate the exponent on the left hand side.
Take the square root of both sides:
\[x+1 = 2~~ or ~~ x+1 = -2\]Look at the first equation: Solve for x.
Subtract 1 from both sides:
\[x = 1~~ or ~~ x+1 = -2 \]Look at the second equation: Solve for x.
Subtract 1 from both sides
Which leaves you with..?

Answer 2

Didn't I delete it ? My thinking process when wrong. I DO recall that x²+2x+1 is a perfect square trinomial, and I know that 3 is 2 units away from 1, to get to the perfect square trinomial, but I don't get 1, I get -1..... this is the thinking error that I encountered.

not like anyone is interested anyway, though.

Answer 3

ok im getting confused a bit

Answer 4

Oh, don't be afraid to hurt my feelings, I don't think I have them. Well... at least it's not like that I seem to have them :)