Question

ODPONENTS AND POLYNONIALS Solving a word problem using a quadratic equation with rationa... The length of a rectangle is 5 yd more than double the width, and the area of the rectangle is 42 yd^2? Find the dimensions of the rectangle. Length: width:

Answer 1

Start by sketching the rectangle

write the area as an equation

`\begin{gathered} A=L\cdot W \\ 42=(x)(2x+5) \\ 42=2x^2+5x \end{gathered}`

write the expression in the form of a quadratic expression

`ax^2+bx+c=0`
`2x^2+5x-42=0`

solve the equation using the quadratic equation

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

x=-6; x=3.5

since we are talking about meassurements it cannot be a negative, so the solution must be 3.5

Dimensions of the rectangles

Lenght: (2x+5) = 12

Width: 3.5