Question

solve for n.
$n^2 - 8n - 48 = 0$
a. 4 or -12
b. -4 or 12
c. 6 or 8
d. -6 or -8
e. -6 or 8

Explanation:

Step1: Factor the quadratic equation

We need to factor \( n^2 - 8n - 48 = 0 \). We look for two numbers that multiply to \(-48\) and add up to \(-8\). The numbers are \(-12\) and \(4\) because \(-12\times4 = -48\) and \(-12 + 4=-8\). So we can factor the equation as \((n - 12)(n + 4)=0\)? Wait, no, wait. Wait, let's do it correctly. Wait, if we have \(n^2 - 8n - 48\), we need two numbers \(a\) and \(b\) such that \(a\times b=-48\) and \(a + b=-8\). Let's see, \( -12\) and \(4\): \(-12\times4=-48\), \(-12 + 4=-8\). So the factoring is \((n - 12)(n + 4)=0\)? Wait, no, \((n + 4)(n - 12)=n^2-12n + 4n-48=n^2 - 8n - 48\), yes. So then we set each factor equal to zero:

Step2: Solve for n

Set \(n + 4 = 0\), then \(n=-4\). Set \(n - 12 = 0\), then \(n = 12\). So the solutions are \(n=-4\) or \(n = 12\).

Answer:

B. \(-4\) or \(12\)