A quadratic equation with integral coefficient has integral roots. Justify your answer

Hint: We start solving this question by discussing the methods for disproving a statement and check if we can disprove the given statement in the method of disproving by a counter example. So, we take different examples of quadratic equations with integral coefficients and see if it has integral roots. If we get an equation that doesn’t satisfy the given statement then we say that the given statement is false.

Complete step-by-step answer:
We can disprove any statement in one of the three ways,
1) By Counter Example
2) By disproving existence statements
3) Disproof by Contradiction
Let us follow the method one and see if we can disprove the given statement.
The statement we are given is, a quadratic equation with integral coefficient has integral roots.
So, for any quadratic equation $a{{x}^{2}}+bx+c$ with integral coefficients $a,b,c$, roots are also integers.
Let us check if it is true for a few examples.
Let us consider a quadratic equation, say, ${{x}^{2}}-2x+1$.
Now let us find the roots of the above equation.
Let us consider the formula, the roots of any quadratic equation $a{{x}^{2}}+bx+c$ are given by
$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So, using this formula we get the roots of the quadratic equation ${{x}^{2}}-2x+1$ as,
$\begin{align}
  & \Rightarrow x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 1 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{2\pm \sqrt{4-4}}{2} \\
 & \Rightarrow x=\dfrac{2\pm \sqrt{0}}{2} \\
 & \Rightarrow x=\dfrac{2\pm 0}{2} \\
 & \Rightarrow x=\dfrac{2}{2} \\
 & \Rightarrow x=1 \\
\end{align}$
So, the roots are 1 and 1, that is integers.

Let us consider a quadratic equation, say, $4{{x}^{2}}-4x+1$.
Now let us find the roots of the above equation.
Let us consider the formula, the roots of any quadratic equation $a{{x}^{2}}+bx+c$ are given by
$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So, using this formula we get the roots of the quadratic equation $4{{x}^{2}}-4x+1$ as,
$\begin{align}
  & \Rightarrow x=\dfrac{-\left( -4 \right)\pm \sqrt{{{\left( -4 \right)}^{2}}-4\left( 4 \right)\left( 1 \right)}}{2\left( 4 \right)} \\
 & \Rightarrow x=\dfrac{4\pm \sqrt{16-16}}{8} \\
 & \Rightarrow x=\dfrac{4\pm \sqrt{0}}{8} \\
 & \Rightarrow x=\dfrac{4}{8} \\
 & \Rightarrow x=\dfrac{1}{2} \\
\end{align}$
So, both roots of the quadratic equation with integer coefficients, $4{{x}^{2}}-4x+1$ are $\dfrac{1}{2}$, that is roots are not integers.
So, this equation, $4{{x}^{2}}-4x+1$, does not satisfy the given statement.
As discussed above as we have shown a counter example of the given statement, we can say that the given statement is not true.
Hence the answer is False.

Note: The common mistake one makes while solving this question is one might start taking examples which only give integral roots, as we have done in the first example in the solution and mark the answer as true. But it is wrong. So, we need to take different types of possible equations with integral coefficients and check the roots.