Answered

At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

The length of a rectangle is 5 greater than 3 times the width. The area of the rectangle is 22cm. Find the length and the width. (SHOW STEPS)

Sagot :

[tex]x-width\\3x+5-length\\22cm^2-area\ of\ the\ rectangle\\\\x\cdot(3x+5)=22\\3x^2+5x-22=0\\\\a=3;\ b=5;\ c=-22\\\\\Delta=b^2-4ac\\\\\Delta=5^2-4\cdot3\cdot(-22)=25+264=289\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{289}=17[/tex]


[tex]x_1=\frac{-5-17}{2\cdot3}=\frac{-22}{6} < 0\\\\x_2=\frac{-5+17}{2\cdot3}=\frac{12}{6}=2\\\\3x+5=3\cdot2+5=6+5=11\\\\Answer:the\ width\ is\ 2cm\ and\ the\ lenght\ is\ 11cm.[/tex]
kate200468
[tex]a-the\ length\ of\ a\ rectangle,\ \ \ a>0\\b-the\ width\ of\ a\ rectangle,\ \ \ b>0\\\\a=3b+5\ \ \ \wedge\ \ \ A=22\ cm^2\\ \\A=a\cdot b\ \ \ \Leftrightarrow\ \ \ (3b+5)\cdot b=22\ \ \Leftrightarrow \ \ 3b^2+5b-22=0\\ \\\Delta=5^2-4\cdot3\cdot(-22)=25+264=289\ \ \Rightarrow\ \ \sqrt{\Delta} =17\\ \\b_1= \frac{-5-17}{2\cdot3} = \frac{-22}{6}=-3 \frac{2}{3} <0,\ \ \ \ \ \ \ \ \ b_2= \frac{-5+17}{2\cdot3} = \frac{12}{6}=2\\ \\b=2\ \ \ \Rightarrow\ \ \ \ a=3 b+5=3\cdot2+5=6+5=11[/tex]

[tex]Ans.\ the\ length\ is\ 11\ cm,\ \ the\ width\ is\ 2\ cm.[/tex]
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.