First slide

Theory of equations

Question

If a root of the equation ${ax}^{2} + bx + c = 0$ be reciprocal of a root of the equation then $a^{'} x^{2} + b^{'} + c^{'} = 0$ , then

Moderate

A

${({cc}^{'} - {aa}^{'})}^{2} = ({ba}^{1} - {cb}^{1}) ({ab}^{1} - {bc}^{1})$
B

${({bb}^{'} - {aa}^{'})}^{2} = ({ca}^{'} - {bc}^{'}) ({ab}^{'} - {bc}^{1})$
C

${({cc}^{1} - {aa}^{'})}^{2} = ({ba}^{1} + {cb}^{'}) ({ab}^{'} + {bc}^{'})$
D

None of these

Solution

Let $α$ be a root of first equation, then $\frac{1}{α}$ be a root of second equation. Therefore ${aα}^{2} + bα + c = 0$ and $a^{'} \frac{1}{α^{2}} + b^{'} \frac{1}{α} + c^{'} = 0$
or $c^{'} α^{2} + b^{'} α + a^{'} = 0$
Hence $\frac{α^{2}}{{ba}^{'} - b^{'} c} = \frac{α}{{cc}^{'} - {aa}^{'}} = \frac{1}{{ab}^{'} - {bc}^{'}}$
${({cc}^{'} - {aa}^{'})}^{2} = ({ba}^{'} - {cb}^{'}) ({ab}^{'} - {bc}^{'}) .$

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