Question

The dimensions of a rectangle can be given by x+9 and x-2. If the area is 60ft^2, what is the value of X?

Answer 1

Answer:

$x=5$

Dimensions of rectangle are

W=14 and L = 3

Step-by-step explanation:

Are of rectangle is

$A=WL$ --------(1)

Where W is width and L is length of rectangle

Here

A=60

W=x+9

L=x-2

Substituting values in equation (1)

$60 = (x+9) (x-2)$

$60 = x^{2} +9x -2x -18$

$x^{2} + 7x - 78 = 0$

Using quadratic formula

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

Here a=1, b=7 and C=-78

$x= \frac{-7 +/- \sqrt{7^{2} - 4(-78)} }{2}$                   (Here +/- = ±)

$x= \frac{-7 +/- \sqrt{ 49 + 312} }{2}$

$x= \frac{-7 +/- \sqrt{ 361} }{2}\\x= \frac{-7+/- 19}{2}$$x= \frac{-7 + 19 }{2} \ \ and \ \ x= \frac{-7-19 }{2}$

$x= \frac{12 }{2} \ \ and \ \ x= \frac{-26 }{2}$

$x= 5 \ \ and \ \ x= -13$

x cannot be negative (Negative x means -ve dimensions of rectangle). Therefore

$x=5$

Dimensions of the rectangle are

$W=x+9 = 5+9 = 14$

$L= x - 2 = 5-2 = 3$