If the roots of the equation (a2+b2)x2-2(ac+bd)x+(c2+d2)=0 are equal, prove that a/b=c/d

Roots are real only when the discriminant d is positive or 0. 


Since d = -4(ad - bc)^2, then roots are real only when ad - bc = 0, ie, ad = bc. 


Now, using the quadratic formula we get just one root = -(ac + bd)/(a^2 + b^2). 

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You can also check: 


(x + (ac + bd)/(a^2 + b^2))^2 = x^2 + 2(ac + bd)/(a^2 + b^2)x + (ac + bd)^2/(a^2 + b^2)^2 and note 


(ac + bd)^2 = a^2c^2 + 2abcd + b^2d^2 = a^2c^2 + abcd + abcd + b^2d^2 = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 = (a^2 + b^2)(c^2 + d^2). 


So x^2 + 2(ac + bd)/(a^2 + b^2)x + (ac + bd)^2/(a^2 + b^2)^2 = x^2 + 2(ac + bd)/(a^2 + b^2)x + (c^2 + d^2)/(a^2 + b^2)

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5/ 1/3=15 i think sorry if im wrong