Chapter 9.5, Problem 29PPS

To determine

Find the solution.

Answer to Problem 29PPS

  $x=-10.33$ or $x=5.33$

Explanation of Solution

Given Information: Area of Rectangle = $110f{t}^2$ , Width w = $2x$ ft., Length l = $x+5$ ft.

Formula Used : Area of Rectangle = $Lenght\times Width$ , $a^2+2ab+b^2={(a+b)}^2$ and

  $a^2-b^2=(a+b)(a-b)$

Calculation:

Area of Rectangle = $110f{t}^2$ , Width w = $2x$ ft., Length l = $x+5$ ft.

Area of Rectangle = $Lenght\times Width$

  $110=\left(x+5\right)\times 2x$

So,

  $110=2x^2+10x$

Now, simplify

  $2x^2+10x-110=0$

Now, to make coefficient of $x^2$ = 1, divide the equation by 2.

  $x^2+5x-55=0$

Now, adding & subtracting the square of half the coefficient of x.

  $x^2+5x-55+{\left(\frac{5}{2}\right)}^2-{\left(\frac{5}{2}\right)}^2=0$

Now use formula, $a^2+2ab+b^2={(a+b)}^2$

  ${\left(x+\frac{5}{2}\right)}^2-55-{\left(\frac{5}{2}\right)}^2=0$

So,

  ${\left(x+\frac{5}{2}\right)}^2-55-\frac{25}{4}=0$

So,

  ${\left(x+\frac{5}{2}\right)}^2-\left(\frac{220+25}{4}\right)=0$

Now,

  ${\left(x+\frac{5}{2}\right)}^2-\left(\frac{245}{4}\right)=0$

Now, square root of 245 is $\sqrt{245}$ and square root of 4 is 2

  ${\left(x+\frac{5}{2}\right)}^2-{\left(\frac{\sqrt{245}}{2}\right)}^2=0$

Now use formula, $a^2-b^2=(a+b)(a-b)$

  $\left(x+\frac{5}{2}+\frac{\sqrt{245}}{2}\right)\left(x+\frac{5}{2}-\frac{\sqrt{245}}{2}\right)=0$

Now,

  $\left(x+\frac{5+\sqrt{245}}{2}\right)\left(x+\frac{5-\sqrt{245}}{2}\right)=0$

Now,

  $\left(x+10.33\right)\left(x-5.33\right)=0$

Hence,

  $x=-10.33$ or $x=5.33$