If `x^2+p x+q=0a n dx^2+q x+p=0,(p!=q)` have a common roots, show that `1+p+q=0` . Also, show that their other roots are the roots of the equation `x^2+x+p q=0.`

The roots of the equation (q-r)x^(2)+(r-p)x+(p-q)=0

If p and q are roots of the equation 2x^2-7x+3 then p^2+q^2=

The equations x^3+4x^2+p x+q=0 and x^3+6x^2+p x+r=0 have two common roots, where p,q,r in R. If their uncommon roots are the roots of equation x^2+2a x+8c=0, then (i) a+c=8 (ii) a+c=2 (iii) 3q=2r (iv) 2q=3r