Question

If the equations (1+^m2) x^2 + 2mcx + c^2a^2=0 has equal roots. Show that c^2=a^2(1+m2)

Answer 1

CORRECT QUESTION :–

• If the equations (1 + m²) x² + 2mcx + m²a² = 0 has equal roots. Show that c² = a²(1+m²).

ANSWER :–

GIVEN :–

• The equations (1 + m²) x² + 2mcx + m²a² = 0 has equal roots.

TO PROVE :–

=> c² = a²(1+m²)

SOLUTION :–

• If a quadratic equation ax² + bx + c = 0 has equal roots , then Discriminant will be zero.

$\\ \: \: \: \longrightarrow \: \: \large { \boxed{ \bold{ D = 0}}}$

• And we know that Discriminant –

$\\ \: \: \: \longrightarrow \: \: \large { \boxed{ \bold{ D = {b}^{2} - 4ac}}}$

• Here –

$\\ \: \: \: { \huge{.}} \: \: \: \large { \bold{a = 1 + {m}^{2} }}$

$\\ \: \: \: { \huge{.}} \: \: \: \large { \bold{b = 2mc }}$

$\\ \: \: \: { \huge{.}} \: \: \: \large { \bold{c = {m}^{2} {a}^{2} }}$

• So that –

$\\ \: \: \implies \: \: { \bold{ {b}^{2} - 4ac = 0}}$

• Put the values –

$\\ \: \: \implies \: \: { \bold{ {(2mc)}^{2} - 4(1 + {m}^{2})( {m}^{2} {a}^{2}) = 0}}$

$\\ \: \: \implies \: \: { \bold{ {(2mc)}^{2} = 4(1 + {m}^{2})( {m}^{2} {a}^{2}) }}$

$\\ \: \: \implies \: \: { \bold{ 4 {m}^{2} {c}^{2} = 4(1 + {m}^{2})( {m}^{2} {a}^{2}) }}$

$\\ \: \: \implies \: \: { \bold{ \cancel{ 4 {m}^{2} } {c}^{2} = \cancel4(1 + {m}^{2})( \cancel{{m}^{2}} {a}^{2}) }}$

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

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

$\\ \: \: \longrightarrow \: \: { \bold{ \underline{\underline{ Hence \: \: proved }}}}$

Answer 2

Given Equation :-

• $\sf{(1 + {m}^{2}) {x}^{2} + 2mcx + {m}^{2} {a}^{2} }$

• Equation has equal roots.

To Prove :-

• $\sf{ {c}^{2} = {a}^{2}(1 + {m}^{2}) }$

Solution :-

We know that general form of quadratic equation is :-

$\sf{\implies a {x}^{2} + bx + c = 0 }$

Comparing given equation with general form of quadratic equation :-

→ a = ( 1+m²)

→ b = 2mc

→ c = m²a²

If roots are equal it means that

→ Discreminant (D) = 0

$\sf{\implies {b}^{2} - 4ac = 0 }$

$\sf{ \implies \: ( {2mc)}^{2} - 4(1 + {m}^{2} )( {c}^{2} - {a}^{2}) = 0 }$

$\sf{ \implies \: 4 {m}^{2}. {c}^{2} - 4( {c}^{2} - {a}^{2}+ {m}^{2}{c}^{2} - {m}^{2}. {a}^{2}= 0 }$

$\sf{ \implies \: 4[ {m}^{2} . {c}^{2} - ({c}^{2} - {a}^{2}+ {m}^{2}{c}^{2} - {m}^{2}. {a}^{2})]= 0 }$

$\sf{ \implies \: {m}^{2} . {c}^{2} - {c}^{2} + {a}^{2}- {m}^{2}{c}^{2} + {m}^{2}. {a}^{2}= 0 }$

$\sf{ \implies \:- {c}^{2} + {a}^{2}+ {m}^{2}. {a}^{2}= 0 }$

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

Hence proved.