Chapter 11.1, Problem 50SPR

To determine

To find: The value of $x$ to the nearest tenth.

Answer to Problem 50SPR

The value of $x=4$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

Use the secant product theorem as,

  $\begin{array}{l}x\left(x+5\right)=3\left(3+9\right)\\ x^2+5x=36\\ x^2+5x-36=0\end{array}$

Then, use the quadratic formula to determine the roots of the equation as,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Then,

  $\begin{array}{c}x=\frac{-\left(5\right)\pm \sqrt{5^2-4\left(1\right)\left(-36\right)}}{2\left(1\right)}\\ =\frac{-5\pm \sqrt{25+4\left(36\right)}}{2}\\ =\frac{-5\pm \sqrt{13}}{2}\\ =-9,4\end{array}$

The, side length must be positive, thus the value of $x=4$ .