Question

If each pair among the equations $x^2+qr=0$, $x^2+qx+rp=0$ and $x^2+rx+pq=0$ have a common root then the product of these common roots is

• A. $-p{q}^2$
• B. $pq(p+q)$
• C. $-(p{q}^2+p{q}^2)$
• D. $2pqr$

Answer 1

The correct option is D $2pqr$

$x^2+px+qr=0$and$x^2+qx+rp=0$ have common root a,

$x^2+qx+rp=0$and$x^2+rx+pq=0$ have common root b,

$x^2+rx+pq=0$and$x^2+px+qr=0$ have common root c,

From the first assumption,a will satisfy both the equations,

$a^2+pa+qr=0$and$a^2+qa+rp=0$

subtract these two equations,

$(a^2+pa+qr)-(a^2+qa+rp)=0-0$

$pa+qr-qa-rp=0$

$pa-qa+qr-rp=0$

$a(p-q)-r(p-q)=0$

$(a-r)(p-q)=0$

$a-r=0;p-q=0$

$a=r,p=q\dots .\left(i\right)$

These c and assumption,b will satisfy both these equations,

$b^2+qb+rp=0$and$b^2+rb+pq=0$

Subtracting these two,

$b(q-r)-p(q-r)=0$

$(b-p)(q-r)=0$

$b=p,q=r\dots .\left(ii\right)$

The third assumption,c will satisfy both these equations,

$c^2+rc+pq=0$and$c^2+pc+qr=0$

Subtracting these two,

$c(r-p)-q(r-p)=0$

$(c-q)(r-p)=0$

$c=q,r=p\dots .\left(iii\right)$

Now from equations(i)(ii)and(iii) we see that $p=q,q=r,r=p,$

So we get the common roots $a=r,b=p,c=q$ and their product $=2abc=2pqr$