M x belongs to R. compare the square of X - x + 1 with the square of - 2m - 2mx | Math enthusiast | Mathematical art

M x belongs to R. compare the square of X - x + 1 with the square of - 2m - 2mx

X ^ 2-x + 1 - (- 2m ^ 2-2mx) = x ^ 2 + (2m-1) x + 2m ^ 2 + 1 make it > 0 constant △ = (2m-1) ^ 2-4 (2m ^ 2 + 1) = 4m ^ 2 + 1-4m-8m ^ 2-4 = - 4m ^ 2-4m-30 this discriminant △ = 16-16 * 30 is constant, so the discriminant of x ^ 2 + (2m-1) x + 2m ^ 2 + 1 is less than 0 constant. So x ^ 2-x + 1 - (- 2m ^ 2-2mx) > 0 is constant, so x ^ 2-x + 1 > - 2m ^ 2-2mx is constant, so x ^ 2-x + 1 > - 2m ^ 2-2mx is constant

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