For `a`, `b`,`c` non-zero, real distinct, the equation, `(a^(2)+b^(2))x^(2)-2b(a+c)x+b^(2)+c^(2)=0` has non-zero real roots. One of these roots is also the root of the equation :

Options

A. `(b^(2)-c^(2))x^(2)+2a(b-c)x-a^(2)=0`
B. `(b^(2)+c^(2))x^(2)-2a(b+c)x+a^(2)=0`
C. `a^(2)x^(2)+a(c-b)x-bc=0`
D. `a^(2)x^(2)-a(b-c)x+bc=0`

Answer

C

Explanation

`(c )` ` (a^(2)+b^(2))x^(2)-2b(a+c)x+b^(2)+c^(2)=0`
`D=4b^(2)(a+c)^(2)-4(a^(2)+b^(2))(b^(2)+c^(2))`
`=-4(b^(4)-2b^(2)ac+a^(2)c^(2))`
`=-4(b^(2)-ac)^(2)`
For real roots, `D ge 0`
`implies -4(b^(2)-ac)^(2) ge 0`
`implies b^(2)-ac=0`
`implies` Roots are real and equal.
`:.` Roots are `(2b(a+c))/(2(a^(2)+b^(2)))`
`=(b(a+c))/(a^(2)+ac)`
`=(b)/(a)`
This root satisfies option `(c )`