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form the equation whose roots are the squares of the sum and of the difference of the roots of 2x²+2(m+n)x+m²+n²=0.

THUDURI ANUSHA , 12 Years ago
Grade
anser 2 Answers
Rinkoo Gupta
sum of the roots =-(m+n)
product of roots=(m^2+n^2)/2
difference of roots= square of sum of roots - 4. product of roots
(m+n)^2-4(m^2+n^2)/2
= - (m-n)^2
The sum of roots of new eq =(m+n)^2 - (m-n)^2=4mn
product of roots=(m+n)^2.{ -(m-n)^2}=-(m^2-n^2)
Hence New eq is x^2-4mn.x -(m^2-n^2)^2=0

Thanks & Regards
Rinkoo Gupta
AskIITians Faculty
Last Activity: 11 Years ago
Sher Mohammad
x1+x2=sum of roots of given equation= -(m+n)
x1x2=product of roots=m^2+n^2/2
(x1-x2)^2= (x1+x2)^2-4x1x2=(m+n)^2-2(m^2+n^2)= 2mn-m^2-n^2=-(m-n)^2
the equation is x^2-(m+n)^2x-(m-n)^2
Last Activity: 11 Years ago
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