If the roots of the given equation ({m^2} + 1 | vedclass

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Question

(C) For a quadratic equation $Ax^2 + Bx + C = 0$,if the roots are equal,the discriminant $D = B^2 - 4AC$ must be equal to $0$.Here,$A = ({m^2} + 1)$,$

Vedclass · Accepted Answer

${a^2} - {b^2}({m^2} + 1) = 0$

Vedclass · Answer

(C) For a quadratic equation $Ax^2 + Bx + C = 0$,if the roots are equal,the discriminant $D = B^2 - 4AC$ must be equal to $0$.Here,$A = ({m^2} + 1)$,$B = 2am$,and $C = ({a^2} - {b^2})$.Setting $D = 0$:$(2am)^2 - 4({m^2} + 1)({a^2} - {b^2}) = 0$$4a^2m^2 - 4({m^2}a^2 - {m^2}b^2 + a^2 - b^2) = 0$Dividing by $4$:$a^2m^2 - (a^2m^2 - m^2b^2 + a^2 - b^2) = 0$$a^2m^2 - a^2m^2 + m^2b^2 - a^2 + b^2 = 0$$m^2b^2 + b^2 - a^2 = 0$$b^2(m^2 + 1) - a^2 = 0$Therefore,${a^2} - {b^2}({m^2} + 1) = 0$.

If the roots of the given equation $({m^2} + 1){x^2} + 2amx + {a^2} - {b^2} = 0$ are equal,then

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