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AKMaths
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#41
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Yes it’s correct. Remember from national 5 that triangles in a semi circle are right angled at the circumference. As we know this is a right angled triangle, as tangent to radius makes a right angle, then the side of the triangle opposite the right angle is the diameter of the circle. Question like this in old paper.
(Original post by hazellowe)
Why assume diameter? Do you think thats right?
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AKMaths
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#42
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(x-1) is a factor but the quotient does not factorise; the discriminant is needed to show there are no real roots. The discriminant from the quotient bracket is ‘-8’ so impossible to find roots or factorise.
(Original post by Aaron_0610)
I think for paper 2Question 10 (b), you can actually factorise further into a 4 sets of bracket answer. I found a second factor (x-1) but then I successfully factorised what was in the third bracket into two further brackets. So you get something like but not exactly (x+3)(x-1)(3x - a)(x+b) --- i can't remember the 2 values I got for a and b, I don't think discriminant had anything to do with this question, but I could be wrong.
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Aaron_0610
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#43
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(Original post by AKMaths)
(x-1) is a factor but the quotient does not factorise; the discriminant is needed to show there are no real roots. The discriminant from the quotient bracket is ‘-8’ so impossible to find roots or factorise.
Oh well, despite that still added up what I got and projecting 80-82%
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