Question
If the roots of the equation $${x^2} - 2ax + {a^2} + a - 3 = 0$$ are less than 3 then
A.
$$a < 2$$
B.
$$2 \leqslant a \leqslant 3$$
C.
$$3 < a \leqslant 4$$
D.
$$a > 4$$
Answer :
$$a < 2$$
Solution :
$$\eqalign{
& \alpha - 3 < 0,\beta - 3 < 0\,\,{\text{and }}D \geqslant 0 \cr
& \Rightarrow \,\,\alpha + \beta - 6 < 0,\alpha \beta - 3\left( {\alpha + \beta } \right) + 9 > 0\,\,{\text{and }}4{a^2} - 4\left( {{a^2} + a - 3} \right) \geqslant 0 \cr
& \Rightarrow \,\,2a - 6 < 0,{a^2} + a - 3 - 3 \cdot 2a + 9 > 0\,\,{\text{and }}3 - a \geqslant 0. \cr} $$
Solve these and find the common value.