Question

Statement-I : If $a+b+c>0$ and $a<0<b<c$, then the roots of the equation $a(x-b)(x-c)+b(x-c)(x-a)+c(x-a)(x-b)=0$ are of both negative.
Statement-II : If both roots are negative, then sum of roots $<0$ and product of roots $>0$.

• A. Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I
• B. Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I
• C. Statement-I is true, Statement-II is false
• D. Statement-I is false, Statement-II is true

Answer 1

The correct option is D Statement-I is false, Statement-II is true

Statement-I : If $a+b+c>0anda<0<b<c$, then the roots of the equation $a(x-b)(x-c)+b(x-c)(x-a)+c(x-a)(x-b)=0$ are of both negative.
$a(x-b)(x-c)+b(x-c)(x-a)+c(x-a)(x-b)=(a+b+c)x^2-2(ab+bc+ca)x+3abc=0$
let $\alpha ,\beta$ are roots of the equation.
$\alpha \beta =\frac{3abc}{a+b+c}<0$ as $a+b+c>0$ and $a<0<b<c$
$\therefore \alpha \beta <0$
$\Rightarrow$ Roots are of opposite sign.
Statement-I is false, Statement-II is true
Hence, option D.