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

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

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

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(C) For a quadratic equation $Ax^2 + Bx + C = 0$,the roots are equal if the discriminant $D = B^2 - 4AC = 0$.Here,$A = (m^2 + 1)$,$B = 2am$,and $C = a^2 - b^2$.Substituting these values into the condition $B^2 - 4AC = 0$:$(2am)^2 - 4(m^2 + 1)(a^2 - b^2) = 0$$4a^2m^2 - 4(m^2a^2 - m^2b^2 + a^2 - b^2) = 0$Dividing by $4$:$a^2m^2 - (m^2a^2 - m^2b^2 + a^2 - b^2) = 0$$a^2m^2 - m^2a^2 + m^2b^2 - a^2 + b^2 = 0$$m^2b^2 - a^2 + b^2 = 0$$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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