Question

If the roots of $x^2-ax+b=0$ are two consecutive odd integers, then $a^2-4b$ is

• A. 3
• B. 4
• C. 5
• D. 6
• E. 7

Answer 1

Answer: (B)

Let two consecutive odd integers are $(\alpha +1)$and $(\alpha +3)$
$\therefore$ Sum of roots, $\alpha +1+\alpha +3=+a$
$\Rightarrow 2\alpha =a-4$
$\Rightarrow \alpha =\frac{a-4}{2}$
and product of roots, $(\alpha +1)(\alpha +3)=b$
$\Rightarrow {\alpha }^2+4\alpha +3=b$
$\Rightarrow {\left(\frac{a-4}{2}\right)}^2+4\left(\frac{a-4}{2}\right)+3=b$
$\Rightarrow \frac{a^2+16-8a}{4}+\frac{4a-16}{2}+3=b$
$\Rightarrow a^2+16-8a+8a-32+12=4b$
$\Rightarrow a^2-4b=4$