Use complete the square to solve x^2+2x+5=0
To do this, I am going to use a method slightly different than is often taught. I am going to re-write the equation like this -
(x^2+2x+ ___)+5=0 now what number can I put in the blank that will make the expression in the parenthesis a perfect square? it has to be a number that is half of b (in ax^2+bx+c) and is the square of that number. b =2 and half of 2 is 1, 1 squared is 1 so the number I can substitute into the blank would be 1. Now I have added 1 to the left side of the equation, to keep it equal I have to either also subtract 1 from the left side, or add 1 to the right side of the equation. I choose to subtract it from the left side like this -
(x^2+2x+1)+5-1=0 Now what is in the parenthesis is a perfect square trinomial and it factors to (x+1)(x+1) or (x+1)^2 and my new equation simplified becomes
(x+1)^2+4=0 subtract 4 from both sides, and you get
(x+1)^2=(-4) take the square root of both sides to get
x+1=+/-sqrt(-4) re-write the right side to the product of sqrt(-1)*sqrt(4) which simplifies to (sqrt(-1) = i and sqrt(4)=2)=+/-2i this then gives you
x+1=+/-2i subtract 1 from both sides and the result is
x=+/-2i-1 which means that there are no real zero solutions to the equation, convention has us writing imaginary numbers in the form a+bi so we would re write our answer like this
x=-1+/-2i or
x=-1+2i, -1-2i
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