Question

If the roots of the equation $x^2+px+q=0$ differ from the roots of the equation $x^2+qx+p=0$ by the same quantity, then what is the value of (p+q)?

• A. 4
• B. -4
• C. 8

Answer 1

The correct option is B -4

Let $x_1,x_2$ be the roots of the equation $x^2+px+q=0$
and $x_3,x_4$ be the roots of the equation $x^2+qx+p=0$.
Hence, $x_1+x_2=-p,x_1\times x_2=q,x_3+x_4=-q,x_3\times x_4=p$ ...(1)
According to the question, $x_1-x_2=x_3-x_4$
or, $(x_1-x_2{)}^2=(x_3-x_4{)}^{2}or(x_1+x_2{)}^2-4x_1\times x_2=(x_3+x_4{)}^2-4x_3\times x_4$ ...(2)
Putting the values from (1) in (2), we obtain $(p-q)(p+q+4)=0$
Hence, either p = q (not possible otherwise both the equations will become same) or p + q = -4