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secondary 3 | E Maths
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Date Posted: 1 year ago
Views: 206
by completing the square
(a)
-x²+6x+5 = 0
x²-6x = 5
(x-3)² = 5+3² = 14
x = 3±√14 (Ans)
(b)
2x²-8x+14 = 0
x²-4x+7 = 0
x²-4x = -7
(x-2)² = 2²-7 = -3
x = 2±√(-3) (Ans)
(a)
-x²+6x+5 = 0
x²-6x = 5
(x-3)² = 5+3² = 14
x = 3±√14 (Ans)
(b)
2x²-8x+14 = 0
x²-4x+7 = 0
x²-4x = -7
(x-2)² = 2²-7 = -3
x = 2±√(-3) (Ans)
So is there no solution? -3 cannot be square root?
-3 is negative
so the answer will be complex numbers
so the answer will be complex numbers
Therefore need to reject and hence, no real solutions at all. Our secondary school syllabus covers only real numbers. We don’t study complex numbers at secondary school level.
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