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

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

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

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

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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