Write x^2 + 3x + 1 in the form of (x+b)^2 + c

This question is asking for a complete the square method. This is where you take the coefficient of x (the number before x, so 3 here) and divide it by two, this gives you b in the new form. However, upon squaring out this bracket, you will get a number which is b^2 = 9/4 , but we have +1 at the end of the initial equation, so we need to take the necessary amount away from the equation to give us +1 when the second equation is expanded out. This number will give us c. To calculate this we take 1 away from 9/4, and use the negative of that number as c. So: x^2 + 3x + 1 b = 3/2 (x + 3/2)^2 = x^2 + 3x + 9/4 9/4 - 1 = 5/4

so our answer is: (x+3/2)^2 - 5/4