Question
If $$m$$ and $$n$$ are the roots of the equation $$\left( {x + p} \right)\left( {x + q} \right) - k = 0,$$ then the roots of the equation $$\left( {x - m} \right)\left( {x - n} \right) + k = 0,$$ are
A.
$$p$$ and $$q$$
B.
$$\frac{1}{p}{\text{and}}\frac{1}{q}$$
C.
$$- p$$ and $$- q$$
D.
$$p + q$$ and $$p - q$$
Answer :
$$- p$$ and $$- q$$
Solution :
Here, $$m$$ and $$n$$ are the roots of equation.
$$\eqalign{
& \left( {x + p} \right)\left( {x + q} \right) - k = 0 \cr
& {x^2} + x\left( {p + q} \right) + pq - k = 0\,\,\,\,.....\left( {\text{i}} \right) \cr} $$
If $$m$$ and $$n$$ are the roots of equation, then
$$\eqalign{
& \left( {x - m} \right)\left( {x - n} \right) = 0 \cr
& \therefore {x^2} - \left( {m + n} \right)x + mn = 0\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Now equation (i) should be equal to equation (ii),
$$\left( {m + n} \right) = - \left( {p + q} \right)\,\,{\text{and }}mn = pq - k$$
Now, we have to find roots of $$\left( {x - m} \right)\left( {x - n} \right) + k = 0$$
$$\eqalign{
& {x^2} - \left( {m + n} \right)x + mn + k = 0 \cr
& {x^2} + \left( {p + q} \right)x + \left( {pq - k} \right) + k = 0 \cr
& {x^2} + \left( {p + q} \right)x + pq = 0 \cr
& {x^2} + px + qx + pq = 0 \cr
& x\left( {x + p} \right) + q\left( {x + p} \right) = 0 \cr
& \therefore x + q = 0\,\,{\text{or }}x + p = 0 \cr
& \therefore x = - q\,\,{\text{and }}x = - p \cr} $$