Question

If the roots of x^2 -bx/ax-c = k-1/k+1 are numerically equal and opposite in sign then k=?​

Answer 1

$\textbf{Given:}$

$\dfrac{x^2-bx}{ax-c}=\dfrac{k-1}{k+1}$

$\textbf{to find:}$

$\text{The value of k if the roots are numerically equal and opposite in sign}$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{x^2-bx}{ax-c}=\dfrac{k-1}{k+1}$

$x^2-bx=(\dfrac{k-1}{k+1})(ax-c)$

$x^2-bx=a(\dfrac{k-1}{k+1})x-c(\dfrac{k-1}{k+1})$

$x^2-(b+a(\dfrac{k-1}{k+1}))x+c(\dfrac{k-1}{k+1})=0$

$\text{Since the roots are numerically equal and opposite in sign, we have}$

$\textbf{Sum of the roots}=0$

$\dfrac{-\text{ocoefficient of x }}{\text{ocoefficient of x^2 }}=0$

$\implies\,b+a(\dfrac{k-1}{k+1})=0$

$\implies\,b(k+1)+a(k-1)=0$

$\implies\,bk+b+ak-a=0$

$\implies\,k(a+b)=a-b$

$\implies\bf\,k=\dfrac{a-b}{a+b}$

$\therefore\textbf{The value of k is \bf\dfrac{a-b}{a+b} }$

Answer 2

Given:

$\bold{\frac{x^2 -bx}{ax-c} =\frac{k-1}{k+1}}$

To find:

k=?

Solution:

$\to \frac{x^2 -bx}{ax-c} =\frac{k-1}{k+1} .....(i)$

cross multiplication of the above equation:

$\to (x^2 -bx)(k+1) =(k-1)(ax-c)\\\\\to kx^2 +x^2-bkx-bx =akx-ck -ax +c\\\\\to kx^2 +x^2-bkx-bx -akx+ck +ax -c=0\\\\\to x^2(k +1)- bkx-bx -akx+ax +ck-c=0\\\\\to x^2(k +1)-x( bk+b+ak-a) +ck-c=0$

compare the value with the standard equation:

$\to ax^2+bx+c=0$

$\to a= k+1\\\\\to b= -(bk+b+ak-a)\\\\\to c= ck-c$

In the equation root is equal but in opposite sign then:

if $\alpha$ is root then:

$\to \alpha +(- \alpha)= \frac{-b}{a}$

$\to \alpha -\alpha =\frac{-(-(bk+b+ak-a))}{k+1}\\\\\to 0 =\frac{(bk+b+ak-a)}{k+1}\\\\\to (bk+b+ak-a)=0\\\\\to bk+ak= a-b\\\\\to k(a+b)= a-b\\\\\to \boxed{k= \frac{(a-b)}{(a+b)}}$