Question

If the roots of the equation $a{x}^2+bx+a_1^2+b_1^2+c_1^2-a_1{b}_1-a_1{c}_1-b_1{c}_1=0$ are non real then

• A. $2(a-b)+\sum (a_1-b_1{)}^2>0$
• B. $2(b-a)+\sum (a_1+b_1{)}^2=0$
• C. $2(b-a)+\sum (a_1+b_1{)}^2<0$
• D. $2(a-b)+\sum (a_1+b_1{)}^2=0$

Answer 1

The correct option is A

$2(a-b)+\sum (a_1-b_1{)}^2>0$

$Letf\left(x\right)=a{x}^2+bx+a_1^2+b_1^2+c_1^2-a_1{b}_1-a_1{c}_1-b_1{c}_1$
Given non - real roots $\Rightarrow$ f(x) will have same sign $\forall x$.
$\therefore f\left(0\right)=a_1^2+b_1^2+c_1^2-a_1{b}_1-b_1{c}_1-c_1{a}_1=\frac{1}{2}(2a_1^2+2b_1^2+2c_1^2-2a_1{b}_1-2b_1{c}_1-2c_1{a}_1)=\frac{1}{2}\left[\right(a_1-b_1{)}^2+(b_1-c_1{)}^2+(c_1-a_1{)}^2]>0\therefore f(-1)>0\Rightarrow a-b+a_1^2+b_1^2+c_1^2-a_1{b}_1-a_1{c}_1-b_1{c}_1>0\Rightarrow \frac{2(a-b)+(a_1-b_1{)}^2+(b_1-c_1{)}^2+(c_1-a_1{)}^2}{2}>0$