Answer:
[tex]x_1 =-14.43 in, x_2 = 2.426in[/tex]
Since the measurement can't be negative the correct answer for this case would be [tex] x =2.426 in[/tex]
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:
[tex] A_i = 7 in *5 in = 35in^2[/tex]
And we know that the initial area is a half of the entire area in red [tex] A_i = \frac{A_f}{2}[/tex], so then:
[tex] A_f = 2 A_ i =2*35 = 70[/tex]
And we know that the area for a rectangular pieces is the length multiplied by the width so we have this:
[tex] 70 = (x+7) (x+5) [/tex]
We multiply both terms using algebra and the distributive property and we got:
[tex] 70 =x^2 +12 x +5[/tex]
And we can rewrite the expression like this:
[tex] x^2 +12 x -35 = 0[/tex]
And we can solve this using the quadratic formula given by:
[tex] x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex]
Where [tex] a = 1, b =12 and c=-35[/tex] if we replace we got:
[tex] x = \frac{-12 \pm \sqrt{(12)^2 -4(1)(-35)}}{2*1}[/tex]
And the two possible solutions are then:
[tex]x_1 =-14.43 in, x_2 = 2.426in[/tex]
Since the measurement can't be negative the correct answer for this case would be [tex] x =2.426 in[/tex]
