Sometimes. It depends on the values given to the variables.
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a.
That a = ±b
(b + 2c)(b - c)
If we assume that the sqare root of 5 is a rational number, then we can write it as a/b in its simplest form, where a and b have no common factors. Therefore 5 = a2/b2 Therefore 5b2 = a2 Therefore a2 is divisible by 5, because b2 is an integer Therefore a is divisible by 5, because 5 is a prime number. Therefore 5 = 5c/b, where c is an integer Therefore 1 = c/b Therefore c = b Therefore sqrt 5 = 5c/c = 5, which is impossible. So sqrt5 cannot be expressed in the form a/b, and is irrational.
a^2 + b^2 + c^2 - ab - bc - ca = 0=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0 => a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca + a^2 = 0 => (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 Each term on the left hand side is a square and so it is non-negative. Since their sum is zero, each term must be zero. Therefore: a - b = 0 => a = b b - c = 0 => b = c.