If `alpha` and `beta` be two real roots of the equation `x^3+px^2 + qx r = 0`,`(r!=0)` satisfying the real `alpha beta+1=0`then prove that `r^2 + pr + q + 1 =0`.

If `alpha` and `beta` be two real roots of the equation `x^3+px^2 + qx

1.	If `alpha` and `beta` be two real roots of the equation `x^3+px^2 + qx r = 0`,`(r!=0)` satisfying the real `alpha beta+1=0`then prove that `r^2 + pr + q + 1 =0`.
Answer» `x^3+px^2+qx+r=0` has 3 roots `alpha,beta and gamma` now product of roots=>`alphabetagamma=-r` given `alphabeta=-1` thus`gamma=r` so -r is a root of the equation so it should satisfy it , so replace x with -r we get, `r(r^2+pr+q+1)=0` but r cannot be 0 so=>`(r^2+pr+q+1)=0`. Hence Proved.

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If `alpha` and `beta` be two real roots of the equation `x^3+px^2 + qx r = 0`,`(r!=0)` satisfying the real `alpha beta+1=0`then prove that `r^2 + pr + q + 1 =0`.

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