Question

Your school wants to take out an ad in the paper congratulating the basketball team on a successful​ season, as shown to the right. The area of the photo will be half the area of the entire ad. What is the value of​ x?

Answer 1

Answer:

1.4in

Step-by-step explanation:

Length of Photo = 4in

Width of Photo = 3in

Unknown:

Value of X = ?

Solution:

Follow these steps:

  Area of a rectangle  = l x w

Since the photo is a rectangle; area of photo:

  Area of photo  = 4in x 3in = 12in²

For the area of the ad;

   Length of ad  = 4 + x

   Width of ad = 3 + x

Given that,

                  the area of the photo  = $\frac{1}{2}$ area of ad

                           12in²  = $\frac{1}{2}$ area of ad

                   Area of ad  = 24in²

Area of the ad;

             (4 + x) (3 + x)  = 24

               12 + 4x + 3x + x² = 24

                12 + 7x + x²  = 24

                       x² + 7x = 24 - 12

                       x² + 7x = 12

                     x² + 7x - 12  = 0

            Using the almighty formula where

     a  = 1, b = 7 and c = -12

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

     

          x = $\frac{-7 + \sqrt[]{-7^{2} - 4x1x-12 } }{2x1}$   or $\frac{-7 - \sqrt[]{-7^{2} - 4x1x-12 } }{2x1}$

          x  = 1.4  or  -8.4

therefore the answer is 1.4in

x is 1.4in