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, $(\alpha+1)(\alpha+3)=b$
$\Rightarrow \alpha^{2}+4 \alpha+3=b$
$\Rightarrow \left(\frac{a-4}{2}\right)^{2}+4\left(\frac{a-4}{2}\right)+3=b$
$\Rightarrow \frac{a^{2}+16-8 a}{4}+\frac{4 a-16}{2}+3=b$
$\Rightarrow a^{2}+16-8 a+8 a-32+12=4 b$
$\Rightarrow a^{2}-4 b=4$