Question

solve it fast pls it is important​

Answer 1

Given

$\bf(1 + {m}^{2} ) {x}^{2} + 2mcx + {c}^{2} - {a}^{2} = 0 \: has \: equal \: roots$

To prove

$\bf \: {c}^{2} = {a}^{2} (1 + {m}^{2} )$

Concept used

$\bf \: d = {b}^{2} - 4ac$

Proof

$\because \bf roots \: are \: real \: and \: equal \\ \therefore \bf d = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\bf \: d = {b}^{2} - 4ac \: \: \: \: \: \: \\ \\ \implies \bf \: 0 = {b}^{2} - 4ac \: or \: {b}^{2} = 4ac \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf \: here \: a = 1 + {m}^{2} \\ \bf \: b = 2mc \\ \bf c = {c}^{2} - {a}^{2} \\ \\ \implies \bf ({2mc})^{2} = 4(1 + {m}^{2} )( {c}^{2} - {a}^{2} ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \bf 4 {m}^{2} {c}^{2} = 4( {c}^{2} - {a}^{2} + {m}^{2} {c}^{2} - {m}^{2} {a}^{2} ) \: \: \: \\ \\ \implies \bf {m}^{2} {c}^{2} = {c}^{2} - {a}^{2} + {m}^{2} {c}^{2} - {m}^{2} {a}^{2} ) \: \: \: \: \: \: \: \: \: \\ \\ \implies \bf 0 = {c}^{2} - {a}^{2} - {m}^{2} {a}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf \implies {c}^{2} = {a}^{2} + {m}^{2} {a}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \underline{ \boxed{ \huge \bf \: {c}^{2} = {a}^{2}(1 + {m}^{2} ) }} \: \: \: \\ \bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{hence \: proved}$

Note

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