Question

1. $7n^2 + 50n + 48 = 0$

Explanation:

Step1: Factor the quadratic equation

We need to factor \(7n^2 + 50n + 48 = 0\). We look for two numbers that multiply to \(7\times48 = 336\) and add up to \(50\). The numbers are \(28\) and \(24\) (since \(28\times12 = 336\)? Wait, no, \(28 + 24 = 52\), no. Wait, \(7n^2 + 50n + 48\). Let's try to factor by grouping.

We rewrite \(50n\) as \(28n + 22n\)? No, wait, \(7\times48 = 336\). Let's find two numbers that multiply to \(336\) and add to \(50\). Let's see, \(336\div 14 = 24\), \(14 + 24 = 38\). No. \(336\div 21 = 16\), \(21 + 16 = 37\). No. Wait, maybe I made a mistake. Let's use the quadratic formula. The quadratic formula is \(n=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\) for \(ax^2+bx+c = 0\). Here, \(a = 7\), \(b = 50\), \(c = 48\).

Step2: Apply the quadratic formula

First, calculate the discriminant \(D = b^2 - 4ac = 50^2 - 4\times7\times48 = 2500 - 1344 = 1156\). Then \(\sqrt{D}=\sqrt{1156}=34\). Then \(n=\frac{-50\pm34}{2\times7}\).

For the first solution: \(n=\frac{-50 + 34}{14}=\frac{-16}{14}=\frac{-8}{7}\).

For the second solution: \(n=\frac{-50 - 34}{14}=\frac{-84}{14}=-6\).

Wait, but maybe factoring is possible. Let's try again. \(7n^2 + 50n + 48\). Let's write it as \(7n^2 + 42n + 8n + 48\) (since \(42 + 8 = 50\) and \(42\times8 = 336\), and \(7\times48 = 336\)). Then factor by grouping:

\(7n(n + 6) + 8(n + 6) = 0\), so \((7n + 8)(n + 6) = 0\).

Then set each factor to zero:

\(7n + 8 = 0\) or \(n + 6 = 0\).

Solving \(7n + 8 = 0\) gives \(n = -\frac{8}{7}\).

Solving \(n + 6 = 0\) gives \(n = -6\).

Answer:

The solutions are \(n = -6\) and \(n = -\frac{8}{7}\)