IIT JEE , AIEEE Forum - Mathematics , Algebra

dixit_se

if the roots "alpha" n "beta" of quadratic equation ax^2+bx+c=0 are real and of opposite sign. then show that roots of the equation:
"alpha"(x-"beta")^2+ "beta(x-"alpha")^2=0 are also real and of opposite sign.

nivedh_89

the equation comes out 2 be x^2(alpha +beta)-4xalphabeta+alpha*beta*(alpha+beta)
= x^2*-b/a+4xc/a-bc/asquare

sum of the roots = 4c/b
product = c/a = same as product of alpha n beta...........
so roots r of opposite signs.........

now,b^2>=4ac..........

applying this we get that determinant of the new equation is also >=0..........
therefore roots r real...!!!!!!!!!!!!!!!!!!!!!

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