Chapter 9.5, Problem 52E

To determine

To determine the value of $x$ .

Answer to Problem 52E

Breadth= $11ft$

Length= $19ft$

Explanation of Solution

Given:

  

Area = $209f{t}^2$ .

Calculation:

According to question,

Area = $209f{t}^2$ .

Formula for area of rectangle $=l\times b$

Breadth= $4x-5$

  $ft$

Length= $4x+3$

  $ft$

Therefore, area of rectangle $=(4x-5)\times (4x+3)$

So,

  $209=(4x-5)\times (4x+3)$

On solving brackets,

  $209=16x^2+12x-20x-15$

So,

  $209=16x^2-8x-15$

Taking all the terms on one side,

  $16x^2-8x-15-209=0$

So,

  $16x^2-8x-224=0$

Finding the value of $x$

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

On comparing the equation with $a{x}^2+bx+c$

  $a=16,b=-8,c=-224$

Substituting the values,

  $x=\frac{-(-8)\pm \sqrt{{(-8)}^2-4(16)(-224)}}{2(16)}$

Opening brackets,

  $x=\frac{8\pm \sqrt{64+14336}}{32}$

So,

  $x=\frac{8\pm \sqrt{14400}}{32}$

Solving root,

  $x=\frac{8\pm 120}{32}$

Therefore,

  $x=\frac{8+120}{32}$ or $x=\frac{8-120}{32}$

So,

  $x=\frac{128}{32}$

and

  $x=\frac{-112}{32}$

Length can never be negative,

Therefore, $x=4$

Breadth= $4x-5$

  $ft$

Length= $4x+3$

  $ft$

Substituting value of $x$

Breadth= $4(4)-5$ = $16-5$

Breadth= $11ft$

Length= $4(4)+3$ = $16+3$

Length= $19ft$