are these right? (solving quadratic equations)
- Thread starter sarahj3
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- Feb 4, 2004
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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.
. . . . .-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.