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If the roots m and n(m+n is not equal to zero) of the quadratic equation ax2+bx+c=0 are real and of opposite sign.Then show that the roots of the equation m(x-n)2 + n(x-m)2=0 are also real and of opposite sign.
ax2+bx+c=0
m+n=-b/a
mn=c/a
solving m(x-n)2 + n(x-m)2=0
m(x2+n2-2xn) + n(x2+m2-2xm)=0
mx2+mn2-2xmn + nx2+nm2-2xmn=0
x2(m+n) -(2mn)x +mn(m+n) = 0
-bx2/a - 2cx/a - bc/a2 =0
-( abx2 + 2acx+bc )=0
abx2 + 2acx+bc=0
let its root be α & β
α + β = -2c/b
αβ = c/a
α + β = 2/(m+n)
αβ = mn
hence proved
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x2(m+n) -(4mn)x +mn(m+n) = 0
-bx2/a - 4cx/a - bc/a2 =0
-( abx2 + 4acx+bc )=0
abx2 + 4acx+bc=0
α + β = -4c/b
α + β = 4/(m+n)
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