Find the Value of k for which the Given Value is a Solution
Question:
Find the value of \(k\) for which the given value is a solution of the equation:
\[ 7x^2+kx-3=0 \]
Given, \[ x=\frac{2}{3} \]
Solution
Since \(x=\frac{2}{3}\) is a solution, it must satisfy the equation.
Substituting \(x=\frac{2}{3}\) into \[ 7x^2+kx-3=0, \] we get
\[ 7\left(\frac{2}{3}\right)^2+k\left(\frac{2}{3}\right)-3=0 \]
\[ 7\cdot\frac{4}{9}+\frac{2k}{3}-3=0 \]
\[ \frac{28}{9}+\frac{2k}{3}-\frac{27}{9}=0 \]
\[ \frac{1}{9}+\frac{2k}{3}=0 \]
Multiplying both sides by 9:
\[ 1+6k=0 \]
\[ 6k=-1 \]
\[ k=-\frac{1}{6} \]
Answer
Therefore, the required value of \(k\) is
\[ \boxed{k=-\frac{1}{6}} \]