the equation pX squared plus 6X plus q has equal roots and that the roots are equal to -3.find value of p and q?
Do you mean
$\displaystyle px^2 + 6x + q = 0$
has roots x = -3?
Look at the discriminant of the quadratic:
$\displaystyle D = 6^2 - 4pq = 36 - 4pq$
If the two roots of a quadratic equation are equal, then we know the discriminant is 0:
$\displaystyle 36 - 4pq = 0 =\implies q = \frac{9}{p}$
So by the quadratic formula, the solution to $\displaystyle px^2 + 6x + q = 0$ will be:
$\displaystyle x = \frac{-6 \pm D^2}{2p} = -3$
But D = 0, so
$\displaystyle \frac{-6}{2p} = -3$
$\displaystyle -6 = -6p$
$\displaystyle p = 1$
Thus $\displaystyle q = \frac{9}{1} = 9$.
-Dan
If $\displaystyle px^2+6x+q$ has one repeated root; $\displaystyle x=-3$
Then $\displaystyle \implies px^2+6x+q=(x+3)(x+3)$
$\displaystyle \implies px^2+6x+q=x^2+6x+9$
Compare co-effs.