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  4. If the roots α, β of the equation px 2+ qx + r =0 are real and of opposite sign (where p , q , r are real coefficient), then the roots of the equation α(x-β)2+β(x-α)2=0 are

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Q. If the roots $\alpha, \beta$ of the equation $px ^2+ qx + r =0$ are real and of opposite sign (where $p , q , r$ are real coefficient), then the roots of the equation $\alpha(x-\beta)^2+\beta(x-\alpha)^2=0$ are

Complex Numbers and Quadratic Equations

Solution:

The equation $\alpha( x -\beta)^2+\beta( x -\alpha)^2=0$ has the product of its roots given by
$\frac{\alpha \beta(\alpha+\beta)}{(\alpha+\beta)}=\alpha \beta<0 \text { (Given) } $
${\left[(\alpha+\beta) x^2-2 x(\alpha \beta+\alpha \beta)+\alpha \beta(\alpha+\beta)=0\right] \text { i.e. }(\alpha+\beta) x^2-4 \alpha \beta x+\alpha \beta(\alpha+\beta)=0}$
Also the discriminant $16 \alpha^2 \beta^2-4 \alpha \beta(\alpha+\beta)^2=-4 \alpha \beta\left[(\alpha+\beta)^2-4 \alpha \beta\right]=-4 \alpha \beta(\alpha-\beta)^2>0$ So, the roots are of opposite sign and are real.