Which shows 3x^2-18x=21 as a perfect square equation? what are the solution(s)?
a. (x-3)^2=0; -3
b. (x-3)^2=16; -1 and 7
c. x^2-6x+9; -3
d. 3x^2-18x-21=0, -1 and 7

b

Step-by-step explanation:

Given

3x² - 18x = 21 ( divide all terms by 3 )

x² - 6x = 7

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- 3)x + 9 = 7 + 9

(x - 3)² = 16 ( take the square root of both sides )

x - 3 = ± = ± 4 ( add 3 to both sides )

x = 3 ± 4, hence

x = 3 - 4 = - 1 and x = 3 + 4 = 7

let x and 400 - x be the two numbers.  

==> 0.80x + 0.85(400 - x) = 400 - 68 from the statement of the problem.  

< ==> 0.80x + 0.85(400 - x) = 400 - 68  

==> 340 - 0.05x = 332  

< ==> -0.05x = -8 ==> x = 160, the first number ==> 400 - x = 240, the second number.