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Find the nature of the roots of x2 - 18x + 81 = 0.?
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Verified answer
x^2 - 18x + 81 = 0
(x - 9)^2 = 0
x = 9
:-
( x - 9 ) ( x - 9 ) = 0
x = 9 (twice)
b^2 - 4ac =
(-18)^2 - 4(1)(81) =
324 - 324 =
0
There is one root with multiplicity 2.
x^2 - 18x + 81 = 0
(x - 9)^2 = 0
x = 9
Two equal, real, rational, positive, natural, whole, integer roots: both 9
x^2 - 18x + 81 = (x - 9)^2,
so there are two real roots, but they are equal to one another.