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Sagot :
Explanation
Let the length be l and width b W, Since the length of the rectangle is 3 m less than twice the width, we will have
[tex]L=2w-3[/tex]Therefore, the area becomes
[tex]\begin{gathered} A=l\times w \\ 65=w(2w-3) \\ 65=2w^2-3w \\ 2w^2-3w-65=0 \\ \mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:} \\ x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \mathrm{For\:}\quad a=2,\:b=-3,\:c=-65 \\ w_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:2\left(-65\right)}}{2\cdot \:2} \\ \mathrm{Separate\:the\:solutions} \\ w_1=\frac{-\left(-3\right)+23}{2\cdot \:2},\:w_2=\frac{-\left(-3\right)-23}{2\cdot \:2} \\ \mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:} \\ w=\frac{13}{2},\:w=-5 \end{gathered}[/tex]Logically, the width becomes
Answer: width = 6.5m
Therefore, the length becomes
[tex]l=2w-3=2(6.5)-3=13-3=10[/tex]Answer" lenght =10 m
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