The length of a rectangle is 2m more than the width. The area of the rectangle is 20m^2. Find the dimensions of the rectangle, to the nearest tenth.please explain in detail
Let x be the width of the rectangle.
`width = x`
Then, express in math form the condition "length is 2 more than x".
`l e n g th = 2 + x`
Next, apply the formula for the area of rectangle.
`A = lw`
where A is the area, l is the length, and w is the width.
From here, substitute A=20. Also replace w with x and l with 2+x.
`20 = x(2+x)`
Express the equation in quadratic form `ax^2+bx+c=0` .
Use quadratic equation to solve for x.
`x=[-b+-sqrt(b^2-4ac)]/[2a] = [-2+-sqrt(2^2-4(1)(-20))]/(2*1)=(-2+-sqrt84)/2`
`x=(-2+-2sqrt21)/2 = -1+-sqrt21`
`x_1=-1+sqrt21=3.6 ` and `x_2=-1-sqrt21=-5.6`
Since x is the width of the rectangle, take only the positive value. Then, substitute this to determine the length.
`width=x=3.6` , `l e n g th = 2 + x = 2+3.6 = 5.6`
Hence, the width of the rectangle is 3.6m and its length is 5.6m.