Question

If the root 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

• A. $p+q+1=0$
• B. $p+q+2=0$
• C. $p+q+4=0$
• D. None of these

Answer 1

Answer: (C)

$p+q+4=0$

We have,

$x^2+px+q=0......\left(1\right)$

$x^2+qx+p=0.......\left(2\right)$

By equation (1)

Let $a$ and $b$ be the roots of equation (1)

Then,

Sum of roots $a+b=\frac{-p}{1}=-p$

Product $a.b=\frac{q}{1}=q$

According to given question,

Difference of roots

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

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

$a-b=\sqrt{p^2-4q}$

Now, by equation (2)

Let $c$ and $d$ be the roots of equation (2)

Then,

Sum of roots $c+d=\frac{-q}{1}=-q$

Product $a.b=\frac{p}{1}=p$

According to given question,

Difference of roots

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

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

$c-d=\sqrt{p^2-4q}$

Again, according to given question,

Difference of roots of both equations are equal

Therefore,

$a-b=c-d$

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

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

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

$(q-p)(q+p)=4(p-q)$

$(q-p)(p+q)=-4$

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

$q=pandp+q+4=0$

Hence, this is the answer.