Q.9.20 - Intermediate Algebra

Question

Solve $16n^2+40n+25=4$

Step-by-Step Solution

Answer

The solutions of the quadratic equation:  $n=-\frac{3}{4}, n=-\frac{7}{4}$.

Step 1. Given information

Consider the quadratic equation

$16n^2+40n+25$

Factor the perfect square trinomial. 

${(4n+5)}^2=4$

Use the Square Root Property. 

$$\begin{gathered} 4n+5=\pm \sqrt{4} \\ 4n+5=\pm 2 \end{gathered}$$

Subtract 5 from both sides.

$4n=\pm 2-5$

Solve for  $n$.

$n=\frac{\pm 2-5}{4}$

Rewrite to show two solutions. 

$$\begin{gathered} n=\frac{2-5}{4}, n=\frac{-2-5}{4} \\ n=-\frac{3}{4}, n=\frac{-7}{4} \end{gathered}$$

Step 2. Check for solutions.

Substitute $n=-\frac{3}{4}$in the original equation.

$$\begin{gathered} 16n^2+40n+25=4 \\ 16{\left(-\frac{3}{4}\right)}^2+40\left(-\frac{3}{4}\right)+25\stackrel{?}{=}4 \\ 16\left(\frac{9}{16}\right)-30+25\stackrel{?}{=}4 \\ 4=4 \end{gathered}$$

Substitute $n=-\frac{7}{4}$in the original equation.

$$\begin{gathered} 16n^2+40n+25=4 \\ 16{\left(-\frac{7}{4}\right)}^2+40\left(-\frac{7}{4}\right)+25\stackrel{?}{=}4 \\ 16\left(\frac{49}{16}\right)-70+25\stackrel{?}{=}4 \\ 4=4 \end{gathered}$$