Respuesta :
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)
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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).
---------------------------------------...
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)
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
