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Prove that the roots of the equation `(a^4+b^4)x^2+4a b c dx+(c^4+d^4)=0`cannot be different, if real. |
Answer» The discriminant of the given equation is `D= 16a^(2)b^(2)c^(2)d^(2) = (a^(4)+b^(4))(c^(4)+d^(4))` `=-4[(a^(4)+b^(4))(c^(4)+d^(4)) -4 = a^(2)b^(2)c^(3)d^(2)]` `=-4[a^(4)c^(4)+a^(4)d^(4)+ b^(4)c^(4)+b^(4)d^(4)-a^(2)b^(2)c^(2)d^(2)]` `=-4[(a^(4)c^(4)+b^(4)d^(4) -2 a^(2)b^(2)c^(2)d^(2)) + (a^(4)d^(4)+b^(4)c^(4) -2 a^(2)b^(2)c^(2)d^2)] ` `=-4[(a^(2)c^(2)-b^(2)d^(2))^(2) + (a^(2)d^(2)-b^(2)c^(2))^(2)]` (1) Since rootd of the givrn equation are real, we have `D ge 0` `rArr =-4[(a^(2)c^(2)-b^(2)d^(2))^(2) + (a^(2)d^(2)-b^(2)c^(2))^(2)] ge0` or `(a^(2)c^(2)-b^(2)d^(2))^(2) + (a^(2)d^(2)-b^(2)c^(2))^(2)le 0` or `(a^(2)c^(2)-b^(2)d^(2))^(2) + (a^(2)d^(2)-b^(2)c^(2))^(2)= 0 (2)` ( Since sum of two positive quantities connot be negative) From (1) and (2), we get D = 0. Hence, the roots of the given quadratic equation are not different, if real. |
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