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 equal to $0$.Here,$A = (a^2 + b^2)$,$B =

Vedclass · Accepted Answer

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

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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 equal to $0$.Here,$A = (a^2 + b^2)$,$B = -2(ac + bd)$,and $C = (c^2 + d^2)$.Setting $D = 0$ gives:$[-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$$-a^2d^2 - b^2c^2 + 2abcd = 0$$-(a^2d^2 + b^2c^2 - 2abcd) = 0$$-(ad - bc)^2 = 0$$(ad - bc)^2 = 0$$ad = bc$$\frac{a}{b} = \frac{c}{d}$.

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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