Question

Roots of the quadratic equation $(x^2-4x+3)+\lambda (x^2-6x+8)=0,\lambda \in R$ will be

• A. always real
• B. real only when lis positive
• C. real only when l is negative
• D. always imaginary

Answer 1

The correct option is A always real

$(x^2-4x+3)+\lambda (x^2-6x+8)=0$
$x^2(1+\lambda )-2x(2+3\lambda )+(3+8\lambda )=0$
Discriminant. $D=4(2+3\lambda {)}^2-4(1+\lambda \left)\right(3+8\lambda )$
$D=4({\lambda }^2+\lambda +1)$
if $\lambda \in R$ then $D>0$
so root of given quadratic are always real.