If the roots of the equation (a^2 + b^2)t^2 - | vedclass

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(D) For a quadratic equation $At^2 + Bt + C = 0$ to have equal roots,the discriminant $D = B^2 - 4AC$ must be $0$.Here,$A = (a^2 + b^2)$,$B = -2(ac +

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$\frac{a}{b} = \frac{c}{d}$

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(D) For a quadratic equation $At^2 + Bt + C = 0$ to have equal roots,the discriminant $D = B^2 - 4AC$ must be $0$.Here,$A = (a^2 + b^2)$,$B = -2(ac + bd)$,and $C = (c^2 + d^2)$.Setting $B^2 - 4AC = 0$:$[-2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) = 0$$4(a^2c^2 + b^2d^2 + 2abcd) - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0$Dividing by $4$:$(a^2c^2 + b^2d^2 + 2abcd) - (a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0$$2abcd - a^2d^2 - b^2c^2 = 0$$-(a^2d^2 + b^2c^2 - 2abcd) = 0$$-(ad - bc)^2 = 0$$(ad - bc)^2 = 0$$ad = bc$$\frac{a}{b} = \frac{c}{d}$

If the roots of the equation $(a^2 + b^2)t^2 - 2(ac + bd)t + (c^2 + d^2) = 0$ are equal,then:

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