Question

if two roots of the equation
$(1 + m {}^{2} )x {}^{2} + 2mcx + (c {}^{2} - a {}^{2} ) = 0$
are equal, then which of the following option is correct
$a. \: a {}^{2} = c {}^{2} (1 + m {}^{2} )$
$b. \: m {}^{2} = a {}^{2} (1 + c {}^{2} )$
$c. \: c { }^{2} = a {}^{2} (1 + m {}^{2} )$
$d. \: am = cm$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\rm :\longmapsto\:(1 + {m}^{2}) {x}^{2} + 2mcx + ( {c}^{2} - {a}^{2}) = 0$

Further, given that equation have real and equal roots.

We know,

Equation ax² + bx + c = 0, have real and equal roots iff Discriminant, D = b² - 4ac = 0.

So, on comparing the given quadratic equation with ax² + bx + c = 0, we get

$\rm :\longmapsto\:a = 1 + {m}^{2}$

$\rm :\longmapsto\:b = 2mc$

$\rm :\longmapsto\:c = {c}^{2} - {a}^{2}$

As equation have real and equal roots, So

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

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

$\rm :\longmapsto\: {4m}^{2} {c}^{2} - 4({c}^{2} - {a}^{2} + {m}^{2} {c}^{2} - {m}^{2} {a}^{2} ) = 0$

$\rm :\longmapsto\: {4m}^{2} {c}^{2} - 4{c}^{2} + 4{a}^{2} - 4{m}^{2} {c}^{2} + 4{m}^{2} {a}^{2} = 0$

$\rm :\longmapsto\:- 4{c}^{2} + 4{a}^{2} + 4{m}^{2} {a}^{2} = 0$

$\rm :\longmapsto\:4{a}^{2} + 4{m}^{2} {a}^{2} = {4c}^{2}$

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

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

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

Hence,

• Option (c) is correct

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Concept Used :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac