If ratio of the roots of …

Let the roots of the given equation be {tex}\alpha {/tex} and {tex}\beta{/tex}.

Then, {tex}\alpha {/tex} + {tex}\beta{/tex} = {tex}{{ - q} \over p}{/tex}         and {tex}\alpha .\beta = {q \over p}{/tex}

Given:     {tex}{\alpha \over \beta } = {a \over b}{/tex}

Now, {tex}\sqrt {{a \over b}} + \sqrt {{b \over a}} + \sqrt {{q \over p}} = 0{/tex}

=>          {tex}\sqrt {{\alpha \over \beta }} + \sqrt {{\beta \over \alpha }} + \sqrt {\alpha \beta } = 0{/tex}          [Since, {tex}{\alpha \over \beta } = {a \over b}{/tex} and {tex}\alpha .\beta = {q \over p}{/tex}]

=>          {tex}{{\sqrt \alpha } \over {\sqrt \beta }} + {{\sqrt \beta } \over {\sqrt \alpha }} + \sqrt {\alpha \beta } = 0{/tex}

=>          {tex}{{\alpha + \beta + \alpha \beta } \over {\sqrt {\alpha \beta } }} = 0{/tex}

=>          {tex}\alpha + \beta + \alpha \beta = 0{/tex}

=>          {tex}{{ - q} \over p} + {q \over p} = 0{/tex}                            [Since, {tex}\alpha + \beta = {{ - q} \over p}{/tex} and {tex}\alpha \beta = {q \over p}{/tex}]

=>          {tex}-q+q=0{/tex}

=>          0 = 0

Hence proved.