Question

Let $a$ and $c$ be odd prime numbers and $b$ be an integer. If the quadratic equation $a{x}^2+bx+c=0$ has rational roots, then which of the following statement(s) is/are correct?

• A. Roots of the equation are $-1,-\frac{c}{a}.$
• B. Both the roots are independent of the coefficients.
• C. Roots of the equation are $-1,-2.$
• D. One of the root is independent of the coefficients

Answer 1

Answer: (A), (D)

One of the root is independent of the coefficients

$a{x}^2+bx+c=0$

For the quadratic equation to have rational roots, discriminant should be a perfect square,
$b^2-4ac=n^2,n\in Z\Rightarrow (b-n)(b+n)=4ac$

Now,
Case $I:b-n=4a$ and $b+n=c$
$2b=4a+c$

As $c$ is an odd number, so $4a+c$ is odd but $2b$ is even.
So, they cannot be equal.

Case $II:b-n=4c$ and $b+n=a$
$2b=4c+a$

As $a$ is an odd number, so $4c+a$ is odd but $2b$ is even.
So, they cannot be equal.

Case $III:b-n=2a$ and $b+n=2c$
$\Rightarrow 2b=2(a+c)\Rightarrow b=a+c$

Now, assuming the roots are $\alpha ,\beta$
$\alpha +\beta =-\frac{b}{a}$
$=-\frac{a+c}{a}=-1-\frac{c}{a}$
$\alpha \beta =\frac{c}{a}$
$\Rightarrow \alpha =-1$ and $\beta =-\frac{c}{a}$