Question

Find the value of k such that
$\frac{5}{2}$
is root of the quadratic equation
$14 {x}^{2} - 27x + k = 0$

Answer 1

$\underline{\mathfrak{\huge{The\:Question\:asked:}}}$

Find the value of k, such that $\tt{\frac{5}{2}}$ is the root of the quadratic equation : $\tt{14x^{2} - 27x + k = 0}$.

$\underline{\mathfrak{\huge{Here's \:Your\:Answer:}}}$

The procedure : Put the value of x as $\sf{\frac{5}{2}}$ in the given quadratic equation.

Doing so, we get :

$\tt{14(\frac{5^{2}}{2^{2}}) - 27(\frac{5}{2}) + k = 0}$

Solve it further

=》 $\tt{\frac{175}{2} - \frac{135}{2} = (-k)}$

Some more steps to go

=》 $\tt{\frac{175 - 135}{2} = (-k)}$

Last one step

=》 $\tt{(-20) = k}$

The Result :

Thus, the value of k should be = ( -20 ), so that the given quadratic equation has its root equal to $\sf{\frac{5}{2}}$.