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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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) -(4mn)x +mn(m+n) = 0

 -bx²/a - 4cx/a - bc/a² =0

 -( abx² + 4acx+bc )=0

      abx² + 4acx+bc=0

  let its root be α & β

  α + β = -4c/b

       αβ = c/a

   α + β = 4/(m+n)

      αβ  = mn

  hence proved

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