\[ \text { For what value of } \mathrm{k} \text {, does the equation } 7 x^2+\mathbf{1 4 x}+k \mathrm{k} \text { become perfect square? } \]

To find the value of \( k \) such that the quadratic equation \( 7x^2 + 14x + k \) becomes a perfect square, we can use the formula for the square of a binomial: \( (ax + b)^2 = a^2x^2 + 2abx + b^2 \).

Comparing the given equation \( 7x^2 + 14x + k \) with \( (ax + b)^2 \), we see that \( a = \sqrt{7}x \) and \( 2ab = 14x \). Solving for \( b \), we get:

\[
2ab = 2\sqrt{7}x \cdot b = 14x \Rightarrow b = \frac{14x}{2\sqrt{7}x} = \frac{7}{\sqrt{7}} = \sqrt{7}
\]

So, the perfect square trinomial is \( ( \sqrt{7}x + \sqrt{7})^2 \). Therefore, for the given quadratic equation to be a perfect square, \( k = \sqrt{7} \times \sqrt{7} = 7 \).