Using a quadratic equation according to the area of the rectangular mat, it is found that:
- At the top and bottom of the painting, the width is 0.654 inches.
- At the left and right of the painting, the width is of 1.308 inches.
The area of a rectangle of length l and width w is given by:
[tex]A = lw[/tex]
In this problem:
- The length of the figure is of [tex]l = 25 + 4x[/tex].
- The width of the figure is of [tex]w = 21 + 2x[/tex].
- Area of 714 square inches, thus:
[tex]lw = 714[/tex]
[tex](25 + 4x)(21 + 2x) = 714[/tex]
[tex]8x^2 + 134x + 525 = 714[/tex]
[tex]8x^2 + 134x - 189 = 0[/tex]
Which is a quadratic equation with [tex]a = 8, b = 134, c = -189[/tex].
The, to find the widths of the mat, we have to first find the value of x, solving the quadratic equation.
[tex]\Delta = 134^2 - 4(8)(-189) = 24004
[/tex]
[tex]x_{1} = \frac{-134 + \sqrt{24004}}{2(8)} = 0.654[/tex]
[tex]x_{2} = \frac{-134 - \sqrt{24004}}{2(8)} = -9.03[/tex]
We are interest in the positive value of x, so x = 0.654.
- At the top and bottom of the painting, the width is of 0.654 inches.
[tex]2(0.654) = 1.308[/tex]
- At the left and right of the painting, the width is of 1.308 inches.
A similar problem is given at https://brainly.com/question/14099239
