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secondary 4 | E Maths
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Date Posted: 4 years ago
Views: 254
a) x^2 - 8x + 3 = (x - 4)^2 - 13 (completing the square)
b) x^2 - 8x + 3 = 2
(x-4)^2 - 13 = 2
(x-4)^2 = 15
(x-4) = +/- sqrt(15)
x = 4 +/- sqrt(15)
where +/- means plus or minus and sqrt(15) means square root of 15. Just work it out on the calculator.
b) x^2 - 8x + 3 = 2
(x-4)^2 - 13 = 2
(x-4)^2 = 15
(x-4) = +/- sqrt(15)
x = 4 +/- sqrt(15)
where +/- means plus or minus and sqrt(15) means square root of 15. Just work it out on the calculator.
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To complete the square, we perform the following.
1. Check whether the coefficient of x2 is 1 or not. If it's not 1, we must factor out the coefficient of x2 from the entire equation so that the new coefficient of x2 becomes 1. Here, it's already 1 so this step can be skipped.
2. Introduce a term equal to the square of half the coefficient of x, then return it immediately after. Here, the coefficient of x is -8. Half of it is -4 and its square is 16, so we must introduce 16 into the equation and return the 16 immediately after. The introduced 16 is meant to complete the square, while the returned 16 is to be evaluated together with the other constant numbers there.
1. Check whether the coefficient of x2 is 1 or not. If it's not 1, we must factor out the coefficient of x2 from the entire equation so that the new coefficient of x2 becomes 1. Here, it's already 1 so this step can be skipped.
2. Introduce a term equal to the square of half the coefficient of x, then return it immediately after. Here, the coefficient of x is -8. Half of it is -4 and its square is 16, so we must introduce 16 into the equation and return the 16 immediately after. The introduced 16 is meant to complete the square, while the returned 16 is to be evaluated together with the other constant numbers there.
Date Posted:
4 years ago
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