
If the given quadratic equation ${x^2} - bx + c = 0$ has equal integral roots, then
A. b and c are integers
B. b and c are even integers
C. b is an even integer and c is a perfect square of a positive integer
D. None of these
Answer
616.8k+ views
Hint: In order to solve this question, we will assume the roots of the quadratic equation as variables and then by using the property of roots of quadratic equations, we will proceed further.
Complete Step-by-Step solution:
Give equation is ${x^2} - bx + c = 0$
Let $\alpha ,\alpha $ be equal integral roots of the given equation
We know that if the quadratic equation is represented by $p{x^2} - qx + r = 0$ then the relation between their roots can be represented as
Sum of roots = $\dfrac{{ - q}}{p}$
And product of roots $ = \dfrac{r}{p}$
So, here our equation is ${x^2} - bx + c = 0$ it means
\[p = 1,q = - b{\text{ and }}r = c\]
Therefore, sum of roots
$
= \alpha + \alpha = \dfrac{{ - ( - b)}}{1} = b \\
\Rightarrow 2\alpha = b \\
$
And product of the roots
$
= \alpha \times \alpha = \dfrac{c}{1} = c \\
\Rightarrow {\alpha ^2} = c \\
$
Therefore, the value of b is $2\alpha ,$ which represents that b is an even integer ( because $\alpha $ is also an integer.) and the value of c is ${\alpha ^2},$ which represents that c is a perfect square of an integer.
Hence, the correct options are “A” and “C”.
Note: In order to solve these types of questions, remember the standard form of the quadratic equation. Also remember the few properties of the roots of quadratic equations. Such as the root of a quadratic equation always satisfies its equation. A quadratic equation can be formed using the sum of and product of its roots. Complex and surd roots of a quadratic equation always exist in a pair.
Complete Step-by-Step solution:
Give equation is ${x^2} - bx + c = 0$
Let $\alpha ,\alpha $ be equal integral roots of the given equation
We know that if the quadratic equation is represented by $p{x^2} - qx + r = 0$ then the relation between their roots can be represented as
Sum of roots = $\dfrac{{ - q}}{p}$
And product of roots $ = \dfrac{r}{p}$
So, here our equation is ${x^2} - bx + c = 0$ it means
\[p = 1,q = - b{\text{ and }}r = c\]
Therefore, sum of roots
$
= \alpha + \alpha = \dfrac{{ - ( - b)}}{1} = b \\
\Rightarrow 2\alpha = b \\
$
And product of the roots
$
= \alpha \times \alpha = \dfrac{c}{1} = c \\
\Rightarrow {\alpha ^2} = c \\
$
Therefore, the value of b is $2\alpha ,$ which represents that b is an even integer ( because $\alpha $ is also an integer.) and the value of c is ${\alpha ^2},$ which represents that c is a perfect square of an integer.
Hence, the correct options are “A” and “C”.
Note: In order to solve these types of questions, remember the standard form of the quadratic equation. Also remember the few properties of the roots of quadratic equations. Such as the root of a quadratic equation always satisfies its equation. A quadratic equation can be formed using the sum of and product of its roots. Complex and surd roots of a quadratic equation always exist in a pair.
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