find m so that the roots of the equation X² - bx /ax - C = m - 1 / m+1 may be equal in magnitude and opposite in sign ? answer the questions step by step.....
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Answer:
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\dfrac{x^2-bx}{ax-c}=\dfrac{m-1}{m+1}
ax−c
x
2
−bx
=
m+1
m−1
(m+1)(x^2-bx)=(ax-c)(m-1)(m+1)(x
2
−bx)=(ax−c)(m−1)
(m+1)x^2-b(m+1)x-a(m-1)x+c(m-1)=0(m+1)x
2
−b(m+1)x−a(m−1)x+c(m−1)=0
(m+1)x^2-[b(m+1)+a(m-1)]x+c(m-1)=0(m+1)x
2
−[b(m+1)+a(m−1)]x+c(m−1)=0
\text{Now,}Now,
\text{Sum of the roots=$\dfrac{-b}{a}$}Sum of the roots=
a
−b
\implies\alpha+(-\alpha)=\dfrac{b(m+1)+a(m-1)}{m+1}⟹α+(−α)=
m+1
b(m+1)+a(m−1)
\implies\,0=\dfrac{bm+b+am-a}{m+1}⟹0=
m+1
bm+b+am−a
\implies\,bm+b+am-a=0⟹bm+b+am−a=0
\implies\,bm+am=a-b⟹bm+am=a−b
\implies\,m(a+b)=a-b⟹m(a+b)=a−b
\implies\boxed{\bf\,m=\dfrac{a-b}{a+b}}⟹
m=
a+b
a−b
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