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If alpha and beta be two real roots of t...

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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`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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