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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.

## 2 Answers

9 years ago

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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9 years ago

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/a=0

-( abx+ 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

please approve the answer by clicking below

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