Answer:
The only possible value for x is 7.
Explanation:
Recognize that we are given two different equations for the area, A:
1) A = 63 in^2, and
2) A = x^2 - 16x + 126
These two equations must be equal to each other: A = A
and therefore,
x^2 - 16x + 126 = 63, which becomes x^2 - 16x + 63 = 0 if we subtract 63 from both sides.
Let's solve this using "completing the square." Take half of the coefficient of x (which coefficient is -16), halve it (obtaining -8) and square the result (obtaining 64). Now, between -16x and +63 (see the last equation, above), we write +64 - 64, obtaining:
x^2 - 16x + 64 - 64 + 63, which can be rewritten as:
(x - 8)^2 -1 = 0. Note that this has the form (x - h)^2 + k, and that h is 8 and k is -1. Thus, the associated parabolic graph has its vertex at (h, k), or (8, -1). This graph opens up. Because the vertex is below the x-axis, we know that the graph intersects the x-axis in two places.
Let's find these x values. Rewrite (x - 8)^2 -1 = 0 as (x - 8)^2 = 1. Squaring both sides results in x - 8 = 1, which simplifies to x = 9. This x = 9 is 1 greater than the x-coordinate of the vertex (8). Knowing tht the graph is symmetrical about the vertical line x = 8, we can safely assume that x = 7 is the other solution, as it is 1 unit to the left of x = 8.
Now we must check these possible solutions {7, 9}.
Evaluate the second equation at each 7 and 9 and determine whether the area turns out to be 63 in^2, as it must.
(7)^2 - 16(7) + 126 = 49 + 126 - 112, which in turn is equal to 63. Yes, x = 7 is a possible value for x.
Next: x = 9. (9)^2 - 16(9) + 126 = 81 - 112 + 126 = 95. This does not agree with A = 63 in^2, so we must reject x = 9.
The only possible value for x is 7.