Question

If the equation $x^2-2px+q=0$ has two equal roots, then the equation $(1+y)x^2-2(p+y)x+(q+y)=0$ will have its roots real and distinct only when

• A. $y$ is not negative and $p$ is not unity.
• B. $y$ is negative and $p$ is not unity.
• C. $y$ is negative and $p$ is unity.
• D. None of these.

Answer 1

The correct option is B $y$ is negative and $p$ is not unity.

$x^2-2px+q=0$
As roots are equal,
$D=0\Rightarrow {4p}^2-4q=0\Rightarrow p^2=q$
Now in $(1+y)x^2-2(p+y)x+(q+y)=0$
For real and distinct roots,
$4{(p+y)}^2-4(q+y)(1+y)>0\Rightarrow p^2+y^2-2py-q-y-qy-y^2>0\Rightarrow (-2p-q-1)y+p^2-q>0\Rightarrow -(2p+q+1)y>0\Rightarrow +(p^2+2p+1)y<0.\Rightarrow {(p+1)}^{2}y<0.\Rightarrow y<0$