Question

If ${{x}^{2}}-bx+c=0$ has equal integral roots, then
A. b and c are integers.
B. b and c are even integers.
C. b is even integer and c is a perfect square of an integers.
D. none of these

Answer 1

Hint: We will first start letting $\alpha $ be the equal integer roots. Then we will use the fact that if $a{{x}^{2}}+bx+c$ is a quadratic equation. Then their sum of roots is $\dfrac{-b}{a}$ and the product of roots is $\dfrac{c}{a}$. Then using this we will check the options to find the correct answer.

Complete step-by-step answer:
Now, we have been given a quadratic equation $a{{x}^{2}}+bx+c=0$ and also, it has been given that it has equal roots which are integers.
Now, let us the value of equal integer roots be $\alpha $.
Now, we know that for a quadratic equation $a{{x}^{2}}+bx+c$.
Sum of roots $=\dfrac{-b}{a}$
Product of roots $=\dfrac{c}{a}$
Now, we have the equation as ${{x}^{2}}-bx+c$.
Sum of roots $=\alpha +\alpha =b.........\left( 1 \right)$
Product of roots $=\alpha \times \alpha =\dfrac{c}{a}............\left( 2 \right)$
Now, from (1) we have $2\alpha =b$ and since $\alpha $ is an integer. Therefore, the value of b is an even integer.
Now, from (2) we have ${{\alpha }^{2}}=c$ and since $\alpha $ is an integer. Therefore, c is a perfect square also we can see that both b and c are integers for sure as $b=2\alpha \ and\ c={{\alpha }^{2}}$. Hence, the correct options are (A) and (C).

Note: It is important to note the fact that if $a{{x}^{2}}+bx+c=0$ is a quadratic equation then sum of roots $=\dfrac{-b}{a}$ and product of roots $=\dfrac{c}{a}$. Also, it is important to have we have proved b is an even integer and c is a perfect square by using this property of quadratic equation.