Anonymous

9x^2 - 42x + 49 can be the area of a square. 9x^2 - 42x + 49 = (3 x - 7)^2 = (7 - 3 x)^2 x = 7/3 What would x be, if the area of the square is 64 square meters? 9x^2 - 42x + 49 = 64 9x^2 - 42x - 15 = 0 (3x + 1)(3x - 15) = 0 Solutions: x = -1/3 x = 5

Anonymous

9x^2 - 42x + 49 = (3x - 7)^2 > 0 for all x except x = 7/3. Therefore, 9x^2 - 42x + 49 can be the area of a square.

Anonymous

If you mean 9x^2 -42x +49 = 0 then it can be sides of a square 7/3 by 7/3 with an area of 49/9 square units

Anonymous

9x² - 42x + 49 = [ 3x - 7 ] [ 3x - 7 ] Thus can be the area of a square.

Anonymous

YES!!! It can be the area of a square. First factor (3x - 7)(3x - 7) => (3x - 7)^2 An area of a square has both its sides of the same length e.g. 3 x 3 = 3^2 = 9 In this case the two sides both have a length of '3x - 7'

Anonymous

y = 9x^2 - 42x + 49. Does this mean we can choose any x we like? Choose x=0. Then y=49; this is an area of a square with side 7.