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Algebra Level pending

There was a statement on a page on the book.

"x1x_1 and x2x_2 are the roots of the equation 1x2mx+2=0\square_1 x^2 - mx + \square_2 = 0, where 1{m,n}\square_1 \in \{m, n\} and 2{m,n}\square_2 \in \{m, n\} (with mn>0mn > 0 and x1>x2x_1 > x_2). It is true that x1x2+1x2x1+2mn=0\sqrt{\dfrac{x_1}{x_2}} + \triangle_1\sqrt{\dfrac{x_2}{x_1}} + \triangle_2\sqrt{\dfrac{m}{n}} = 0, where 1{1,1}\triangle_1 \in \{-1, 1\} and 2{1,1}\triangle_2 \in \{-1, 1\}."

Which of the following options is correct?

Option 1: 1=m,2=n,1=1,2=1\square_1 = m, \square_2 = n, \triangle_1 = 1, \triangle_2 = -1

Option 2: 1=m,2=n,1=1,2=1\square_1 = m, \square_2 = n, \triangle_1 = -1, \triangle_2 = 1

Option 3: 1=n,2=n,1=1,2=1\square_1 = n, \square_2 = n, \triangle_1 = 1, \triangle_2 = -1

Option 4: 1=n,2=n,1=1,2=1\square_1 = n, \square_2 = n, \triangle_1 = 1, \triangle_2 = 1

Option 5: 1=n,2=m,1=1,2=1\square_1 = n, \square_2 = m, \triangle_1 = -1, \triangle_2 = 1

Option 6: 1=n,2=m,1=1,2=1\square_1 = n, \square_2 = m, \triangle_1 = 1, \triangle_2 = -1

Option 7: 1=n,2=n,1=1,2=1\square_1 = n, \square_2 = n, \triangle_1 = -1, \triangle_2 = 1

Option 8: 1=n,2=n,1=1,2=1\square_1 = n, \square_2 = n, \triangle_1 = -1, \triangle_2 = -1

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