[tex]area: A = 84 \ m ^2 \\perimeter: \ P = 38 \ m\\length: l = \ ?\\ width : \ w=? \\ \\\begin{cases} A=l\cdot w \\ P=2l+2w \end{cases}\\\\\begin{cases} 84=l\cdot w \\ 38=2l+2w \ \ /:2 \end{cases}[/tex]
[tex]\begin{cases} 84=l\cdot w \\ 19= l+ w \end{cases}\\ \\ \begin{cases} 84=l\cdot w \\ l=19-w \end{cases}\\\\ substitution\\\\84 =l\cdot w \\84=(19-w)w\\84=19w-w^2[/tex]
[tex]w^2-19w+84=0\\\\a=1, \ \ b=-19,\ \ c=84 \\\\\\Delta =b^2-4ac = (-19)^2 -4\cdot1\cdot 84 = 361-336=25 \\ \\w_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{19-\sqrt{25}}{2 }=\frac{ 19-5}{2}=\frac{14}{2}=7[/tex]
[tex]w_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{19+\sqrt{25}}{2 }=\frac{ 19+5}{2}=\frac{24}{2}=12 \\ \\\begin{cases}w_{1}= 7\\ 84 = w_{1}\cdot l_{1} \end{cases}\ \ or \ \ \begin{cases}w_{2}= 12\\84 =w_{2}\cdot l_{2} \end{cases} \\\\\begin{cases}w_{1}= 7\\84=7\cdot l_{1}\ \ /:7 \end{cases}\ \ or \ \ \begin{cases}1= 12\\84=12\cdot l_{2} \ \ / :12 \end{cases}\\ \\ \begin{cases}w_{1}= 7\\ l_{1}= \frac{84}{7} \end{cases}\ \ or \ \ \begin{cases}1= 12\\ l_{2}= \frac{84}{12} \end{cases}[/tex]
[tex]Answer : \ \begin{cases}w_{1}= 7\\ l_{1}= 12 \end{cases}\ \ or \ \ \begin{cases}1= 12\\ l_{2}= 7 \end{cases}[/tex]
