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If x^2+p x+q=0a n dx^2+q x+p=0,(p!=q) ha...

If `x^2+p x+q=0a n dx^2+q x+p=0,(p!=q)` have a common roots, show that `p+q=0` . Also, show that their other roots are the roots of the equation `x^2+x+p q=0.`

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Given equation are
`x^(2) + px + q = 0` (1)
and `x^(2) + qx + p = 0` (2)
Clearly for x = 1, from both equations we get `1+ p + q = 0` . Also for (1), product of roots = q
Therefore, (1) has roots x = 1 and x = q
For (2), product of roots = p
Therefoure, (2) has roots x = 1 and x = p
Thus, the equation having roots p and q is
`x^(2) - (p + q) x + pq = 0`
`therefore x^(2) + x + pq = 0 (as 1 + p + q = 0 )`
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