Question
If the roots of the equation $$x^2 – ax + b = 0$$ are real and differ by a quantity which is less than $$c(c > 0),$$ then $$b$$ lies between
A.
$$\frac{{{a^2} - {c^2}}}{4}{\text{and}}\frac{{{a^2}}}{4}$$
B.
$$\frac{{{a^2} + {c^2}}}{4}{\text{and}}\frac{{{a^2}}}{4}$$
C.
$$\frac{{{a^2} - {c^2}}}{2}{\text{and}}\frac{{{a^2}}}{4}$$
D.
None of these
Answer :
$$\frac{{{a^2} - {c^2}}}{4}{\text{and}}\frac{{{a^2}}}{4}$$
Solution :
Given roots are real and distinct, then $$a^2 - 4b > 0$$
$$ \Rightarrow b < \frac{{{a^2}}}{4}$$
Again $$\alpha$$ and $$\beta$$ differ by a quantity less than $$c\left( {c > 0} \right)$$
$$\eqalign{
& \Rightarrow \left| {\alpha - \beta } \right| < c{\text{ or }}{\left( {\alpha - \beta } \right)^2} < {c^2} \cr
& \Rightarrow {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta < {c^2}{\text{ or }}{a^2} - 4b < {c^2}\,\,{\text{or }}\frac{{{a^2} - {c^2}}}{4} < b \cr
& \Rightarrow \frac{{{a^2} - {c^2}}}{4} < b < \frac{{{a^2}}}{4}{\text{by}}\left( 1 \right){\text{and}}\left( 2 \right). \cr} $$