Question

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

• A. $3$
• B. $4$
• C. $5$
• D. $6$
• E. $7$

Answer 1

The correct option is B

$4$

Step 1: Solve for the values of $a$ and $b$

Given quadratic equation is $x^2-ax+b=0$

Say $\alpha ,\beta$ are the roots of the given quadratic equation

We know that for a quadratic equation $a{x}^2+bx+c=0$:

1. Sum of the roots$=\frac{-b}{a}$
2. Product of the roots $=\frac{c}{a}$

Here,

1. Sum of the roots $\Rightarrow \alpha +\beta =a$
2. Product of the roots $\Rightarrow \alpha \beta =b$

Step 2: Solve for the required value

Given that $\alpha$ and $\beta$ are two consecutive odd integers

$\therefore$Say $\alpha =2n+1,\beta =2n+3$

Now,

$\begin{array}{cccl}\therefore & a^2-4b& =& {\left(\alpha +\beta \right)}^2-4\alpha \beta \\ & & =& {\alpha }^2+{\beta }^2+2\alpha \beta -4\alpha \beta \\ & & =& {\alpha }^2+{\beta }^2-2\alpha \beta \\ & & =& {\left(\alpha -\beta \right)}^2\\ & & =& {\left(2n+1-(2n+3)\right)}^2\\ & & =& {\left(-2\right)}^2\\ \Rightarrow & a^2-4b& =& 4\end{array}$

Hence, the correct option is option (B) i.e. $4$.