Question

the length of a rectangle is 10 more than it’s width. If the area of the rectangle is 52yd^2 , find the appropriate width.

Answer 1

The answers to the Math Nation should be

1.58 seconds

10 inches

3.775 yards

Answer 2

The width is
`(-5+√(77))\ yd`

Explanation:

Let

x-----> the length of rectangle

y-----> the width of rectangle

we know that

The area of rectangle is equal to

`A=xy`

`A=52\ yd^(2)`

so

`52=xy` ------> equation A

`x=y+10` -----> equation B

substitute equation B in equation A

`52=(y+10)y`

`y^(2)+10y-52=0`

Solve the quadratic equation

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`y^(2)+10y-52=0`

so

`a=1\\b=10\\c=-52`

substitute in the formula

`y=\frac{-10(+/-)\sqrt{10^(2)-4(1)(-52)}} {2(1)}`

`y=\frac{-10(+/-)√(308)} {2}`

`y1=(-10(+)2√(77))/(2)` -------> the solution is the positive value

`y2=(-10(-)2√(77))/(2)`

The width is
`(-10(+)2√(77))/(2)\ yd=(-5+√(77))\ yd`