Question

the following quadratic equations if they exist by the method of the completing the square 8x²-22x-21=0

Answer 1

HEY!!!


By factoring, 
8x^2 + 22x - 21 = 0 
x^2 + 22/8x - 21/8 = 0/8 
x^2 + 22/8x - 21/8 = 0 
x^2 + 22/8x + (22/8/2)^2 - 289/64 = 0 
x^2 + 22/8x + (121/64) - 289/64 = 0 
x^2 + 22/8x + (11/8)^2 - (17/8)^2 = 0 
[x^2 + 22/8x + (11/8)^2] - [(17/8)^2] = 0 
(x + 11/8)^2 - (17/8)^2 = 0 
Therefore, [(x + 11/8) + (17/8)] * [ (x+11/8) -(17/8)] = 0 
(x + 28/8) * (x - 6/8) = 0 
(x + 7/2) * (x - 3/4) = 0 
x + 7/2 = 0 or x - 3/4 = 0 
x = -7/2 or x =3/4. 

OR, 

By completing the square (another way of factoring), 
8x^2 + 22x - 21 = 0 
x^2 + 22/8x - 21/8 = 0/8 
x^2 + 22/8x = 21/8 
x^2 + 22/8x +(22/8/2)^2 = 21/8 + (22/8/2)^2 
x^2 + 22/8x + (22/8*1/2)^2 = 21/8 + (22/8* 1/2)^2 
x^2 + 22/8x + (11/8)^2 = 21/8 + (11/8)^2 
(x + 11/8)^2 = 21/8 + 121/64 
(x + 11/8)^ 2 = 168/64 + 121/64 
(x + 11/8)^2 = 289/64 
(x+11/8) = square root of 289/64 
x+11/8 = (+-)17/8 
x = (17/8 - 11/8) or (-17/8-11/8) 
x = 6/8 or x = -28/8 
x = 3/4 or x = - 7/2 
x = -7/2 or x = 3/4.