Question

In each of the following, determine the value of k
for which the given value is the solution of the
equation:
3x^2 + kx + 5 = 0;x=-5/3​

Answer 1

Answer:

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Step-by-step explanation:

$put \: the \: value \: of \: x \: in \: eq. \\ 3 \times { \frac{ - 5}{3} }^{2} + k \times \frac{ - 5}{3} + 5 = 0 \\ \\ = > 25 + \frac{ - 5k}{3} + 5 = 0 \\ = > \frac{75 - 5k + 15}{3} = 0 \\ = > 75 - 5k + 15 = 0 \\ = > - 5k = - 90 \\ = > k = 18$

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Answer 2

given equation

3x²+ Kx +5 = 0

and x= -5/3

put x= -5/3

3× (-5/3)²+k (-5/3) +5 = 0

$3 \times \frac{25}{9} - \frac{5k}{3} + 5 = 0$

$\frac{25}{3} - \frac{5k}{3} + 5 = 0$

$\frac{25 - 5k + 15}{3} = 0$

this implies

$25 - 5k + 15 = 0$

$5k = 40$

$k = 8$

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