contestada

the length of a rectangle is 8cm more than the width and its area is 172cm2 .find the width of the rectabgle

Respuesta :

[tex]l-the\ length\\l+8-the\ width\\A=172\ cm^2\\\\A=lengtf\cdot width\\\\subtitute\\\\l(l+8)=172\\\\l^2+8l=172\ \ \ |subtract\ 172\ from\ both\ sides\\\\l^2+8l-172=0\\a=1;\ b=8;\ c=-172\\\\\Delta=b^2-4ac\\\\\Delta=8^2-4\cdot1\cdot(-172)=64+688=752 > 0\\\\l_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ l_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{752}=\sqrt{16\cdot47}=4\sqrt{47}\\\\l_1=\dfrac{-8-4\sqrt{47}}{2\cdot1}=-4-2\sqrt{47} < 0\\\\l_2=\dfrac{-8+4\sqrt{47}}{2\cdot1}=-4+2\sqrt{47} > 0\\\\The\ width:l+8=-4+2\sqrt{47}+8=4+2\sqrt{47}\\\\Answer:\boxed{length=(2\sqrt{47}-4)cm;\ width=(4+2\sqrt{47})cm}[/tex]
[tex]length\approx9.71cm;\ width\approx17.71cm[/tex]