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 $p+q$

• A. $-4$
• B. $-2$
• C. $-3$
• D. None of these

Answer 1

The correct option is A $-4$

The two equations given are

$x^2+px+q=0$ ----- ( 1 )

$x^2+qx+p=0$ ----- ( 2 )

Let $a,b$ be the roots of equation ( 1 ) and $c,d$ be the roots of equation ( 2 )

$\Rightarrow$ $a+b=-p$ and $c+d=-q$

$\Rightarrow$ $ab=q$ and $cd=p$

Now, difference of roots of ( 1 ) is

$(a-b{)}^2=(a+b{)}^2-4ab$

$\Rightarrow$ $a-b=\sqrt{(a+b{)}^2-4ab}=\sqrt{(-p{)}^2-4q}$

And difference of roots of ( 2 ) is

$(c-d{)}^2=(c+d{)}^2-4cd$

$\Rightarrow$ $c-d=\sqrt{(c+d{)}^2-4cd}=\sqrt{(-q{)}^2-4p}$

Now, according to the question, it is given that the difference of roots for both the equation are same.

$\Rightarrow$ $a-b=c-d$

$\Rightarrow$ $\sqrt{(-p{)}^2-4q}=\sqrt{(-q{)}^2-4p}$

$\Rightarrow$ $(-p{)}^2-4q=(-q{)}^2-4p$

$\Rightarrow$ $p^2-4q=q^2-4p$

$\Rightarrow$ $p^2-q^2=4q-4p$

$\Rightarrow$ $p^2-q^2+4p-4q=0$

$\Rightarrow$ $(p+q)(p-q)+4(p-q)=0$

$\Rightarrow$ $(p-q)(p+q+4)=0$

$\Rightarrow$ $p+q+4=0$

$\therefore$ $p+q=-4$