Question

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)

Answer 1

`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\ \ √(\Delta) =17\\ \\b_1= (-5-17)/(2\cdot3) = (-22)/(6)=-3 (2)/(3) <0,\ \ \ \ \ \ \ \ \ b_2= (-5+17)/(2\cdot3) = (12)/(6)=2\\ \\b=2\ \ \ \Rightarrow\ \ \ \ a=3 b+5=3\cdot2+5=6+5=11`


`Ans.\ the\ length\ is\ 11\ cm,\ \ the\ width\ is\ 2\ cm.`

Answer 2

`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=(-b-\sqrt\Delta)/(2a);\ x_2=(-b+\sqrt\Delta)/(2a)\\\\\sqrt\Delta=√(289)=17`



`x_1=(-5-17)/(2\cdot3)=(-22)/(6) < 0\\\\x_2=(-5+17)/(2\cdot3)=(12)/(6)=2\\\\3x+5=3\cdot2+5=6+5=11\\\\Answer:the\ width\ is\ 2cm\ and\ the\ lenght\ is\ 11cm.`