Question

If the coefficients of the quadratic equation $ a{{x}^{2}}+bx+c=0 $ are odd integers, then the roots of the equation cannot be
(A) Rational number
(B) Irrational number
(C) Imaginary number
(D) We can’t say anything about the roots

Answer 1

Hint: For answering questions of these type we will assume some values for a, b and c and find the value of discriminant $ D={{b}^{2}}-4ac $ and determine the nature of roots by determining the discriminant. We can assume values as $ a=2l+1 $ , $ b=2m+1 $ and $ c=2n+1 $ and then substitute them in $ D={{b}^{2}}-4a $ . We can then assume that roots are rational and try to prove it as right or wrong. For that, we can consider D as $ {{\left( 2p+1 \right)}^{2}} $ and equate it with the above obtained value.

Complete step-by-step answer:
As it is given that the coefficients of the quadratic equation $ a{{x}^{2}}+bx+c=0 $ are odd integers, let us assume that $ a=2l+1 $ , $ b=2m+1 $ and $ c=2n+1 $ .
The nature of roots of any quadratic equation is given by its discriminant. Discriminant is given by $ D={{b}^{2}}-4ac $ .
There are 3 possibilities for D less than 0, greater than 0 and equal to 0. And for D greater than 0 there are 2 more sub possibilities that are for D as perfect squares and not a perfect squares.
(i) For D<0, the roots are imaginary and conjugate pairs.
(ii) For D=0, the roots are rational and equal.
(iii) For D>0 and a perfect square, the roots are rational and unequal.
(iv) For D>0 and not a perfect square, the roots are irrational conjugate pairs.
Let us apply this and find discriminant for the quadratic equation we have it will be $ \begin{align}
  & D={{b}^{2}}-4ac \\
 & \Rightarrow D={{\left( 2m+1 \right)}^{2}}-4\left( 2n+1 \right)\left( 2l+1 \right) \\
\end{align} $
We have
 $ \begin{align}
  & \text{odd}\times \text{odd=odd} \\
 & \text{odd}\times even=even \\
 & even\times even=even \\
\end{align} $
 $ \begin{align}
  & even-even=even \\
 & odd-odd=even \\
 & odd-even=odd \\
\end{align} $
 $ \begin{align}
  & D\text{ will be }{{\left( odd \right)}^{2}}-4\left( odd \right)\left( odd \right) \\
 & \Rightarrow D\text{ will be odd-4}\left( odd \right) \\
 & \Rightarrow D\text{ will be odd-even} \\
 & \Rightarrow \text{D will be odd}\text{.} \\
\end{align} $
Hence the discriminant of the quadratic equation we have will be odd.
Let us assume the roots to be rational and verify it. We know that since the D is not equal to 0 for sure so the roots will not be equal. So, let us assume D as $ {{\left( 2p+1 \right)}^{2}} $ .
Hence, we have to verify $ {{\left( 2p+1 \right)}^{2}}={{\left( 2m+1 \right)}^{2}}-4\left( 2n+1 \right)\left( 2l+1 \right) $ .
 $ \begin{align}
  & {{\left( 2p+1 \right)}^{2}}={{\left( 2m+1 \right)}^{2}}-4\left( 2n+1 \right)\left( 2l+1 \right) \\
 & \Rightarrow 4\left( 2n+1 \right)\left( 2l+1 \right)={{\left( 2m+1 \right)}^{2}}-{{\left( 2p+1 \right)}^{2}} \\
 & \Rightarrow 4\left( 2n+1 \right)\left( 2l+1 \right)=\left( 4{{m}^{2}}+4m+1 \right)-\left( 4{{p}^{2}}+4p+1 \right) \\
 & \Rightarrow 4\left( 2n+1 \right)\left( 2l+1 \right)=4\left( {{m}^{2}}+m-{{p}^{2}}-p \right) \\
 & \Rightarrow \left( 2n+1 \right)\left( 2l+1 \right)={{m}^{2}}+m-{{p}^{2}}-p \\
\end{align} $
 $ \begin{align}
  & \Rightarrow \left( 2n+1 \right)\left( 2l+1 \right)={{m}^{2}}+m-{{p}^{2}}-p+mp-mp \\
 & \Rightarrow \left( 2n+1 \right)\left( 2l+1 \right)={{m}^{2}}+m+mp-{{p}^{2}}-p-mp \\
 & \Rightarrow \left( 2n+1 \right)\left( 2l+1 \right)=m\left( m+1+p \right)-p\left( p+1+m \right) \\
 & \Rightarrow \left( 2n+1 \right)\left( 2l+1 \right)=\left( m-p \right)\left( m+1+p \right) \\
\end{align} $ By adding and subtracting $ mp $ in R.H.S.
Here L.H.S is odd and we R.H.S depends on m, p. We have 4 possibilities for m, p.
(i)When m is odd, p is even.
 $ \left( odd-even \right)\left( odd+even+1 \right)=odd\times even\Rightarrow even $ . Hence, both will never be equal. So it is not possible.
(ii)When p is odd, m is even.
 $ \left( odd-even \right)\left( odd+even+1 \right)=odd\times even\Rightarrow even $ . Hence, both will never be equal. So it is not possible.
(iii)When m, p both are odd.
 $ \left( odd-odd \right)\left( odd+odd+1 \right)=even\times odd\Rightarrow even $ . Hence, both will never be equal. So it is not possible.
(iv)When m, p both are even.
 $ \left( even-even \right)\left( even+even+1 \right)=even\times odd\Rightarrow even $ . Hence, both will never be equal. So it is not possible.
Hence, there are no possibilities for the roots to be rational. So the roots are irrational.

So, the correct answer is “Option B”.

Note: Here while answering questions of these types we should remember the results of the operations performed between odd and even numbers. They can be determined by observing the results of normal calculations. We should never ever confuse and replace odd with even.