College Algebra, Chapter 1, 1.5, Section 1.5, Problem 86

Suppose that the two television monitor has the same height. One has a conventional screen, which is $5 in$ wider than it is high. The other has wider, which is $1.8$ times as wide as it is high. The diagonal measure of the wider screen is $14 in$ more than the diagonal measure of the smaller. What is the height of the screens?

If we let $TV_1$ be the smaller and $TV_2$ has the wider screen, then


$
\begin{equation}
\begin{aligned}

& \text{Given on } TV_1
&& \text{Given on } TV_2
\\
\\
& w_1 = 5 + h
&& w_2 = (1.8) h
\\
\\
& d_1
&& d_2 = 14 + d_1

\end{aligned}
\end{equation}
$















By Pythagorean Theorem,


$
\begin{equation}
\begin{aligned}

d_1^2 =& (5 + h)^2 + h^2
\\
\\
d_1^2 =& 25 + 10h + h^2 + h^2
\\
\\
d_1^2 =& 25 + 10h + 2h^2
\\
\\
d_1 =& \sqrt{2h^2 + 10h + 25}
\\
\\
(14 + d_1)^2 =& (1.8h)^2 + h^2
\\
\\
196 + 28d_1 + d_1^2 =& 3.24h^2 + h^2
\\
\\
196 + 28d_1 + d_1^2 =& 4.24 h^2 \qquad \text{Equation 1}


\end{aligned}
\end{equation}
$


Substitute $d_1$ and $d_1^2$ in Equation 1


$
\begin{equation}
\begin{aligned}

196 + 28 \sqrt{2h^2 + 10h + 25} + 25 10h + 2h^2 =& 4.24h^2
&& \text{Combine like terms}
\\
\\
28 \sqrt{2h^2 + 10h + 25} =& 2.24 h^2 - 10h - 221
&& \text{Square both sides}
\\
\\
784(2h^2 + 10h + 25) =& 5.0176 h^4 - 44.8h^3 - 990.08h^2 + 4520h + 48841
&&
\\
\\
1568h^2 + 7840 h + 19600 =& 5.0176 h^4 - 44.8h^3 - 990.08h^2 + 4520h + 48841
&&
\\
\\
5.0176h^4 - 44.8h^3 - 2558.08h^2 - 3320 h + 29241 =& 0
&&


\end{aligned}
\end{equation}
$








Based from the graph the height that crosses the $x$-axis is at $h \approx 27.90 in$