Question

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

Answer 1

 Length = x + 2  (because it is 2 cm more than x)
Width = 2x - 5   (5 cm lest than 2x)Area = 54 cm2  this is the formula to find the area Length × Width = Area (x + 2)(2x - 5) = 542x2 - x - 10 = 54   (this is you area)

  Subtract 54 on both sides of equation to make the right side zero. 2x2 - x - 64 = 0  then use the quadratic formula x = (-b ± √(b2 - 4ac)) / 2a where:a = 2b = -1c = -64 Plug in these values into the formula. x = (1 ± √(1 - 4(-128))) / 4 x = (1 ± √(513)) / 4 x = (1 ± 22.65) / 4 x = (1 + 22.65) / 4          and          x = (1 - 22.65) / 4 x = 5.91                          and          x = -5.41  Check the validity of the x values by adding them to the length and width.  If the length or width should be a negative value, then that value of x is not acceptable. Now x = 5.91 Length = 5.91 + 2  (positive value.)Width = 2(5.91) - 5  ( positive value.) x = 5.91    If we look at this -- x = -5.41, Both length and width will be negative values. We reject this value of x.  The answer is x = 5.91

Hope I helped and sorry it was really long

Answer 2

Answer:

x = 6

Step-by-step explanation:

Given : The area of a rectangle is 56 cm². The length is 2 cm more than x and the width is 5 cm less than twice x.

We have to find the value of x.

 

Given area of rectangle = 56 cm²

Length  is 2 cm more than x  mathematically written as

L = 2 + x

Width is 5 cm less than twice x  mathematically written as

W = 2x - 5

Thus, we know area of rectangle = length × width

Substitute, we  have,

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

Multiply each term of first bracket with each term of second bracket , we have,

$56=2\cdot \:2x+2\left(-5\right)+x\cdot \:2x+x\left(-5\right)$

Simplify, we have,

$56=2x^2-x-10$

Subtract 56 both sides, we have,

$2x^2-x-10-56=56-56$

Simplify , we get,

$2x^2-x-66=0$

Using quadratic formula for the general equation $ax^2+bx+c=0$ , we have $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=-1,\:c=-66:\quad x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4\cdot \:2\left(-66\right)}}{2\cdot \:2}$

Simplify, we get,

$x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{529}}{2\cdot \:2}$

Also, $\sqrt{529}=23$

We get,

$x_{1,\:2}=\frac{-\left(-1\right)\pm 23}{2\cdot \:2}$

Thus,

$x_{1}=\frac{-\left(-1\right)+23}{2\cdot \:2}\\\\\\ x_{2}=\frac{-\left(-1\right)-23}{2\cdot \:2}$

Simplify, we get,

$x_1=6,\:x_2=-\frac{11}{2}$

Since, x has to be whole number so x = 6