Question 199571


From {{{x^2-8x+16}}} we can see that {{{a=1}}}, {{{b=-8}}}, and {{{c=16}}}



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



{{{D=(-8)^2-4(1)(16)}}} Plug in {{{a=1}}}, {{{b=-8}}}, and {{{c=16}}}



{{{D=64-4(1)(16)}}} Square {{{-8}}} to get {{{64}}}



{{{D=64-64}}} Multiply {{{4(1)(16)}}} to get {{{(4)(16)=64}}}



{{{D=0}}} Subtract {{{64}}} from {{{64}}} to get {{{0}}}



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