Respuesta :
The length and width of rectangle are 18 inches and 6 inches respectively
Solution:
Given that area of rectangle is 108 square inches
Let "L" be the length of rectangle
Let "W" be the width of rectangle
To find: dimenions of rectangle. i.e length and width
Given that length is 12 inches longer then the width
Length = 12 + width
L = 12 + W
The area of rectangle is given as:
[tex]\text {Area of rectangle }=\text { length } \times \text { width }[/tex]
Substituting the given values in above formula,
[tex]\begin{array}{l}{108=L \times W} \\\\ {108=(12+W) \times W} \\\\ {108=12 W+W^{2}} \\\\ {W^{2}+12 W-108=0}\end{array}[/tex]
Let us solve the above quadratic equation using quadratic formula
[tex]\text {For a quadratic equation } a x^{2}+b x+c=0[/tex]
[tex]x = \frac{-b \pm \sqrt{\left(b^{2}-4 a c\right)}}{2 a}[/tex]
[tex]\text {So for } \mathrm{W}^{2}+12 \mathrm{W}-108=0, \text { we get } \mathrm{a}=1 \text { and } \mathrm{b}=12 \text { and } \mathrm{c}=-108[/tex]
Substituting the values in quadratic formula,
[tex]\begin{aligned} W &=\frac{-12 \pm \sqrt{12^{2}-4(1)(-108)}}{2 \times 1} \\\\ W &=\frac{-12 \pm \sqrt{144+432}}{2} \\\\ W &=\frac{-12 \pm \sqrt{576}}{2}=\frac{-12 \pm 24}{2} \\\\ W &=\frac{-12+24}{2} \text { or } W=\frac{-12-24}{2} \\\\ W &=6 \text { or } W=-18 \end{aligned}[/tex]
Since width cannot be negative, ignore negative value
So width of rectangle = 6 inches
L = 12 + W = 12 + 6 = 18 inches
So the length and width of rectangle are 18 inches and 6 inches respectively
