Question

Find the value of k for which x=2 is a solution of the equation. kx^2+2x-3=0

Answer 1

$\bold\huge{SOLUTION}$

x = 2 is a solution

Hence,

We substitute x in the equation,

k(2)² + 2(2)-3 = 0

4k + 4 - 3 = 0

4k + 1 = 0

4k = -1

$k \: = \frac{ - 1}{4}$

Answer 2

$here \: x = 2 \:.so \: putting \: x = 2 \: in \: the \: given \: equation \\ k \times {2 }^{2} + 2 \times 2 - 3 = 0 \\ 4k + 4 - 3 = 0 \\ 4k + 1 = 0 \\ 4k = - 1 \\ k = \frac{ - 1}{4}$

it is the correct answer plzz mark as brainlist..