Question

find the value of k in each of the following quadratic equations for which the given value of x is a root of the given quadratic equation 14 x square - 27 X + K is equal to zero therefore x is equal to 5 by 2​

Answer 1

Answer:

Step-by-step explanation: put the value of x in the equation :

14x^2 -27x + k =0

14(5/2)^2-27(5/2) +k =0

87.5-67.5 + k=0

K =(-20)

If you want to verify the answer then put the value of k in the equation and factorize it , you will get the value or the roots.

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Value\:of\:k=-20}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies 14x^{2} -27x + k = 0 }\\ \\ \tt{:\implies x=\frac{5}{2}} \\\\\red{ \underline \bold{To \: Find : }} \\ \tt{: \implies value \: of \: k = ?}$

• According to given question :

$\tt{ : \implies 14x^{2} -27x + k= 0} \\ \\ \text{Putting \: the \: values\:of\:x} \\ \tt{: \implies 14\times(\frac{5}{2})^{2} - 27\times \frac{5}{2} + k= 0 } \\ \\ \tt{: \implies \: 14\times \frac{25}{4}-\frac{135}{2}+k= 0 } \\ \\ \tt{ : \implies \: \frac{175}{2} -\frac{135}{2}+k = 0 } \\\\ \tt{: \implies \frac{175-135+2k}{2}= 0} \\ \\ \tt{: \implies 40+2k=0} \\\\ \tt{:\implies k=\frac{-40}{2}} \\ \\ \green{\tt{: \implies k = -20}}$