Question 627756
From {{{x^2-16x+4}}} we can see that {{{a=1}}}, {{{b=-16}}}, and {{{c=4}}}



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



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



{{{D=256-4(1)(4)}}} Square {{{-16}}} to get {{{256}}}



{{{D=256-16}}} Multiply {{{4(1)(4)}}} to get {{{(4)(4)=16}}}



{{{D=240}}} Subtract {{{16}}} from {{{256}}} to get {{{240}}}



So the discriminant is {{{D=240}}}



Since the discriminant is greater than zero, this means that there are two real solutions.



Since 240 is NOT a perfect square, these two real solutions are irrational solutions.



So we have 2 real and irrational solutions.


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