Question

If 2 is a root of the quadratic equation ${x^2} - kx + 2 = 0$ , then the value of $k$ is equal to

Answer 1

Hint: Let the roots of the quadratic equation ${x^2} - kx + 2 = 0$ be 2,$\alpha $. It is known that the sum of roots are equal to the negative of coefficient of $x$divided by coefficient of ${x^2}$ and the product of roots are equal to the constant divided by coefficient of ${x^2}$. So 1st the value of $\alpha $ from $2 \times \alpha = 2$ then we will substitute $\alpha $in $2 + \alpha = k$ and ${\text{product of two roots = }}\dfrac{{{\text{constant}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$ get the value of k.

Complete step-by-step answer:
Let the roots of ${x^2} - kx + 2 = 0$ is 2,$\alpha $.
Now, we know that
So, $2 \times \alpha = \dfrac{2}{1}$
$\therefore \alpha = 1$
Now, ${\text{sum of two roots = }}\dfrac{{{\text{ - coefficient of x}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$
Therefore, $2 + \alpha = \dfrac{k}{1}$
Substituting $\alpha = 1$ we get
$\Rightarrow$ $k = 3$

Note: Alternative method
If $f\left( x \right) = {x^2} - kx + 2$ and 2 is the root of $f\left( x \right)$ then $f\left( 2 \right) = 0$
So, ${2^2} - 2k + 2 = 0$
On solving we will get the value of k.
$\Rightarrow$ $k = \dfrac{{{2^2} + 2}}{2} $
$\Rightarrow$ $k = 3$