Question

solve for w
$w^2 + 11w + 28 = 0$
show your work here
$w =$
$w =$

Explanation:

Step1: Factor the quadratic equation

We need to find two numbers that multiply to \(28\) and add up to \(11\). The numbers \(4\) and \(7\) work because \(4\times7 = 28\) and \(4 + 7=11\). So we can factor the quadratic as:
\(w^{2}+11w + 28=(w + 4)(w + 7)=0\)

Step2: Solve for \(w\) using zero - product property

The zero - product property states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
For \((w + 4)(w + 7)=0\), we set each factor equal to zero:

• If \(w+4 = 0\), then \(w=-4\) (by subtracting \(4\) from both sides of the equation \(w + 4=0\)).
• If \(w + 7=0\), then \(w=-7\) (by subtracting \(7\) from both sides of the equation \(w + 7=0\)).

Answer:

\(w=-4\), \(w = - 7\)