Chapter 3.1, Problem 35E

To determine

To calculate: .The rectangle for the value $x$ .

Answer to Problem 35E

The required value of $x$ is $4$

Explanation of Solution

Given Information:

The area of rectangle is $36$ and sides of the rectangle are $x$ and $x+5$ .

Formula Used:

Area of rectangle:

  $Area=l\times b$

Where $l$ is length of rectangle and $b$ breadth of rectangle.

Quadratic formula:

If $a{x}^2+bx+c=0$ then $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where $a,b,c$ are constant and $a\ne 0$ .

Calculation:

Consider the area of rectangle is $36$ .

To find the value of $x$ ; use area of triangle formula.

  $Area=l\times b$ .

Substitute $36$ for $Area$ , $x+5$ for $l$ and $x$ for $b$ in the above equation.

  $36=\left(x+5\right)\times x$

Simplify the above equation.

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

Now, use quadratic formula in the above equation.

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $1$ for $a$ , $5$ for $b$ and $-36$ for $c$ in the above equation.

  $\begin{array}{c}x=\frac{-5\pm \sqrt{5^2-4\left(1\right)\left(-36\right)}}{2\left(1\right)}\\ =\frac{-5\pm \sqrt{25+144}}{2}\\ =\frac{-5\pm \sqrt{169}}{2}\\ =\frac{-5\pm 13}{2}\end{array}$

Simplify the equation.

  $\begin{array}{c}x=\frac{-5\pm 13}{2}\\ =\frac{-5+13}{2},\frac{-5-13}{2}\\ =\frac{8}{2},\frac{-18}{2}\\ =4,-9\end{array}$

Since, $x$ is measurement and it is known that measurement cannot be negative.

Thus, the value of $x$ is $4$ .

Hence, the required real value of $x$ is $4$ .