Three sides of a rectangular field have real lengths √f+6, f2-8, and f+4 respectively. The area of the field is 2f2+2f+8 . What is the area of the field?

Answer 1

Strategy: Break the problem into logical alternative equalities, and solve each set as needed.

Tactics: Since three measures are given for the rectangular sides, two of these measures must match. Thus EITHER f ²- 8 = f + 4 ; f ²- f -12 = 0 ; f = (1 ± (1+48)^0.5)/2 ; f = -3 or 4

OR f ² - 8 = √f + 6 ; solve graphically to f = 4 .

OR √f + 6 = f + 4; substitute variables to solve x^2 - x - 2 = 0 ; x = ½ +- (1+8)^(0.5) /2 ; x = 2 or -1; x^2 = f = 1 or 4.

Now that we have some possible values for f, it's time to check them for consistency!

f = 1 does not satisfy the original equation: √f + 6 = f + 4 , so it is rejected.

The -3 solution is not a valid length, as the √f+6 side is then a complex value.

This leaves f = 4 :

Therefore, sides: 8, 8, 8 respectively (internal consistency with the original problem statement), with area 64 units ².

This problem illuminates the importance of checking your answers vs. the original conditions, whenever you have problems involving quadratics or non-integer powers requiring substitutions of variable: your equations may generate spurious values as well as the true ones.