Applications of Quadratics - Questions
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Question 1
43696
The area of a rectangle is fixed at \(10~\text{m}^2\) and its length is 3 more than its width. Find the dimensions of the rectangle.Â
\begin{align}
&\text{Let } x=\text{width }\rightarrow \text{length}=x+3\\
&\begin{aligned}
\therefore x(x+3)&=10 \\
x^{2}+3x-10&=0 \\
(x+5)(x-2)&=0 \\
\therefore x=-5 \text { or } x&=2 \\
x \neq-5\qquad &
\end{aligned}\\
& \therefore \text{Dimensions are } \\
&\quad \text{ Length } 5\mathrm{~m}, \text{ width }=2\mathrm{~m}
\end{align}
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