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Dear student,
Given, x2 + px + q = 0, where p, q are non-zero constant and the roots of this equation is u and v.

Now we know, for the standard equation ax2 + bx + c =0,
Sum of roots = -(b/a)
Product of roots = (c/a)

Comparing the standard equation with given equation, we get
a= 1, b=p, c =q.
Therefore,
Sum of roots = -p = u+v .. (i)
Product of roots = q = uv ... (ii)

Now, another equation qx2 + px + 1=0 is given and we are required to find out its roots.
Again comparing this equation with the standard equation, we get
a= q, b=p, c =1
Therefore,
Sum of roots = -p/q ...(iii)
Product of roots = 1/q ... (iv)

Now dividing equation (i) by equation (ii), we get,
(u+v)/uv = -p/q .... (v)
Comparing equation (iv) with equation (v),
Sum of roots = (u+v)/uv = 1/u + 1/v .. (vi)

Now, taking reciprocal of equation (ii), we get,
1/uv = 1/q ... (vii)
Comparing equation (vii) with (iv), we get,
Product of roots = 1/uv = (1/u) . (1/v) ...   (viii)

As one can observe in equations (vi) and (viii), 1/u and 1/v are forming the sum and product together for the equation qx2 + px + 1 =0,
Hence its roots are 1/u and 1/v.

I hope you understood the solution!
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