Question

The area is a rectangle is 56 cm. The length is 2 cm more than x and the width is 5 cm less than twice x. Round to the nearest whole number.

​

Answer 1

The solution is
`x=6\ cm`

Explanation:

we know that

The area of rectangle is equal to

`A=LW`

we have

`A=56\ cm^(2)`

so

`56=LW` ------> equation A

`L=x+2` -----> equation B

`W=2x-5` -----> equation C

Substitute equation B and equation C in equation A

`56=(x+2)(2x-5)`

solve for x

`56=(x+2)(2x-5)\\56=2x^(2)-5x+4x-10\\2x^(2)-x-66=0`

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

`2x^(2)-x-66=0`

so

`a=2\\b=-1\\c=-66`

substitute in the formula

`x=\frac{1(+/-)\sqrt{-1^(2)-4(2)(-66)}} {2(2)}`

`x=\frac{1(+/-)√(529)} {4}`

`x1=\frac{1(+)23} {4}=6`

`x2=\frac{1(-)23} {4}=-5.5`

The solution is
`x=6\ cm`

`L=6+2=8\ cm`

`W=2(6)-5=7\ cm`