Respuesta :
EXPLANATION:
Given;
We are given the following information. A rectangle has an area of 208 square inches. The width is 2 more than 3 times the length.
Required;
We are required to find the length and the width.
Step-by-step solution;
If the length is given as l, then the width would be 2 plus 3 times l. That is;
[tex]\begin{gathered} If\text{ } \\ length=l \\ Then \\ width=2+3l \end{gathered}[/tex]Note also that the area of a rectangle is calculated as;
[tex]Area=l\times w[/tex]Given that the area is 208 square inches, we can set up the following information;
[tex]\begin{gathered} Area=l\times w \\ Therefore: \\ 208=l(2+3l) \end{gathered}[/tex]We simplify the parenthesis;
[tex]208=2l+3l^2[/tex]We can move all terms to one side of the equation and we have;
[tex]3l^2+2l-208=0[/tex]Now we have a quadratic equation and we shall solve this by the quadratic equation formula.
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]The variables in this case are;
[tex]a=3,b=2,c=-208[/tex]We can now substitute these into the quadratic formula and solve as follows;
[tex]x=\frac{-2\pm\sqrt{2^2-4(3)(-208)}}{2(3)}[/tex][tex]x=\frac{-2\pm\sqrt{4+2496}}{6}[/tex][tex]x=\frac{-2\pm\sqrt{2500}}{6}[/tex][tex]x=\frac{-2\pm50}{6}[/tex]We can now evaluate separately as follows;
[tex]\begin{gathered} x_1=\frac{-2+50}{6}=\frac{48}{6}=8 \\ Also: \\ x_2=\frac{-2-50}{6}=\frac{-52}{6}=-\frac{26}{3} \end{gathered}[/tex]Therefore, the two values of l will be,
[tex]\begin{gathered} l_1=8in \\ l_2=-\frac{26}{3}in \end{gathered}[/tex]We shall take the positive value only, since we know that the length or width of any plane shape is always a positive value.
Therefore, the length of the rectangle is 8 inches. The width can now be determined as follows;
[tex]w=2+3l[/tex][tex]w=2+3(8)[/tex][tex]w=2+24=26[/tex]Therefore;
ANSWER:
[tex]\begin{gathered} length=8in \\ width=26in \end{gathered}[/tex]