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:

$x_1 =-14.43 in, x_2 = 2.426in$

Since the measurement can't be negative the correct answer for this case would be $x =2.426 in$

Step-by-step explanation:

Let's assume that the figure attached illustrate the situation.

For this case the we know that the original area given by:

$A_i = 7 in *5 in = 35in^2$

And we know that the initial area is a half of the entire area in red $A_i = \frac{A_f}{2}$, so then:

$A_f = 2 A_ i =2*35 = 70$

And we know that the area for a rectangular pieces is the length multiplied by the width so we have this:

$70 = (x+7) (x+5)$

We multiply both terms using algebra and the distributive property and we got:

$70 =x^2 +12 x +5$

And we can rewrite the expression like this:

$x^2 +12 x -35 = 0$

And we can solve this using the quadratic formula given by:

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

Where $a = 1, b =12 and c=-35$ if we replace we got:

$x = \frac{-12 \pm \sqrt{(12)^2 -4(1)(-35)}}{2*1}$

And the two possible solutions are then:

$x_1 =-14.43 in, x_2 = 2.426in$

Since the measurement can't be negative the correct answer for this case would be $x =2.426 in$