Question

If $b$ and $c$ are odd integers, then the equation $x^2+bx+c=0$ has

• A. Two odd roots
• B. Two integer roots, one odd and one even
• C. No integer roots
• D. None of the above

Answer 1

The correct option is C No integer roots

$x^2+bx+c=0$

b,c are odd integers

roots, $x=-b\pm \sqrt{b^2-4c}$

In order to have integer roots i.e., assume roots are integers

$b^2-4c=m^2,mϵz$

$b^2-m^2=4c$

if $b^2-m^2$ is divisible by $2$. Then both b,m are even or both m,b are odd

$\therefore m$ is odd ($\because b$ is odd)

Let $b=2k+1,m=2n+1,k,nϵz$

$4(k^2-n^2)+4(k-n)=4c$

$c=(k-n)(k+n+1)$

if $k-n$ is odd, then $k+n+1$ is even.

(Or) if $k-n$ is even then $k+n+1$ is odd

$\therefore$C will be even

But C cannot be even as it is odd (Given)

$\therefore$ roots of $x^2+bx+c$ are not integers.