Question 150315


From {{{x^2+2x+1}}} we can see that {{{a=1}}}, {{{b=2}}}, and {{{c=1}}}



{{{D=b^2-4ac}}} Start with the discriminant formula



{{{D=(2)^2-4(1)(1)}}} Plug in {{{a=1}}}, {{{b=2}}}, and {{{c=1}}}



{{{D=4-4(1)(1)}}} Square {{{2}}} to get {{{4}}}



{{{D=4-4}}} Multiply {{{4(1)(1)}}} to get {{{(4)(1)=4}}}



{{{D=0}}} Subtract {{{4}}} from {{{4}}} to get {{{0}}}



Since the discriminant is equal to zero, this means that there is one real solution