S sarahj3 New member Joined Apr 4, 2006 Messages 10 May 22, 2006 #1 Solve: -x^2 + 6x - 8 = 0 x^2 - 6x + 8 (x - 4) (x - 2) x1 = 4 x2 = 2 Solve: x^2 = 6x - 5 x^2 - 6x + 5 (x - 5) (x - 1) x1 = 5 x2 = 1
Solve: -x^2 + 6x - 8 = 0 x^2 - 6x + 8 (x - 4) (x - 2) x1 = 4 x2 = 2 Solve: x^2 = 6x - 5 x^2 - 6x + 5 (x - 5) (x - 1) x1 = 5 x2 = 1
stapel Elite Member Joined Feb 4, 2004 Messages 16,543 May 22, 2006 #2 The solution to any "solving" exercise may be checked by plugging it back into the original problem. So: . . . . .-x<sup>2</sup> + 6x - 8 = 0 . . . . .x<sub>1</sub> = 4: . . . . . .-(4)<sup>2</sup> + 6(4) - 8 ?=? 0 . . . . . .-16 + 24 - 8 ?=? 0 . . . . . .24 - 24 = 0 . . . . .x<sub>2</sub> = 2: . . . . . .-(2)<sup>2</sup> + 6(2) - 8 ?=? 0 . . . . . .-4 + 12 - 8 ?=? 0 . . . . . .12 - 12 = 0 So your solutions "check" for the first exercise. (Good work!) Now you check your solutions to the second exercise. Eliz.
The solution to any "solving" exercise may be checked by plugging it back into the original problem. So: . . . . .-x<sup>2</sup> + 6x - 8 = 0 . . . . .x<sub>1</sub> = 4: . . . . . .-(4)<sup>2</sup> + 6(4) - 8 ?=? 0 . . . . . .-16 + 24 - 8 ?=? 0 . . . . . .24 - 24 = 0 . . . . .x<sub>2</sub> = 2: . . . . . .-(2)<sup>2</sup> + 6(2) - 8 ?=? 0 . . . . . .-4 + 12 - 8 ?=? 0 . . . . . .12 - 12 = 0 So your solutions "check" for the first exercise. (Good work!) Now you check your solutions to the second exercise. Eliz.