Question

solve for k.
$4k^2 + 28k + 49 = 0$
write each solution as an integer, proper fraction, or improper fraction in simplest form. if there are multiple solutions, separate them with commas.
$k = $

Explanation:

Step1: Recognize the quadratic form

The equation \(4k^2 + 28k + 49 = 0\) is a quadratic equation. Notice that it might be a perfect square trinomial. Recall that \((a + b)^2=a^2 + 2ab + b^2\). For \(4k^2=(2k)^2\), \(49 = 7^2\), and \(28k=2\times(2k)\times7\). So the left - hand side can be factored as a perfect square.
\(4k^2+28k + 49=(2k + 7)^2\)
So the equation becomes \((2k + 7)^2=0\)

Step2: Solve for k

Set \(2k+7 = 0\) (since a square of a number is zero only when the number itself is zero).
Subtract 7 from both sides: \(2k=-7\)
Divide both sides by 2: \(k =-\frac{7}{2}\)

Answer:

\(-\frac{7}{2}\)