Quadratics - the Australian Mathematical Sciences Institute

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{40} • Quadratics Exercise 5 The discriminant is ∆ = 36k 2 . One real solution occurs when ∆ = 0, which is when k = 0. The quadratic is never negative definite, since ∆ ≥ 0 for all values of k. Exercise 6 Divide the wire into 4x cm for the circle and (20−4x) cm for the square. (The 4 is to make the algebra easier.) The total area is then S = 4x2 π + (5 − x)2 . Since the leading coefficient is positive, the parabola has a minimum. We set dS dx = 0 to find the stationary point, which occurs when x = 5π 20π 4+π . Hence, we use 4+π cm for the circle and 80 4+π cm for the square. Exercise 7 x 2 − 4x − 41 Exercise 8 a 4 7 b − 12 49 Exercise 9 We have (α − β) 2 = (α + β) 2 − 4αβ ( = − b ) 2 4c − a a = b2 − 4ac a 2 . (Note that, if a = 1, then (α − β) 2 is the discriminant of the quadratic. Similarly, we can define the discriminant of a monic cubic with roots α,β,γ to be (α − β) 2 (α − γ) 2 (β − γ) 2 , and so on for higher degree equations.) Exercise 10 The arithmetic mean is 17 6 17± 145 6 !) Exercise 11 and the geometric mean is 2. (Note that the actual roots are Successively substitute x = 1, x = 2, x = 3 and solve the resulting equations for a,b,c to obtain x 2 = 3(x − 1) 2 − 3(x − 2) 2 + (x − 3) 2 .
A guide for teachers – Years 11 and 12 • {41} Exercise 12 Substituting the line into the circle we obtain x 2 − 4x + 4 = 0. The discriminant of this quadratic is 0 and so the line is tangent to the circle at (2,1). Exercise 13 The parabola can be written as (y − 1) 2 = 4(x − 1). So the vertex is (1,1), the focal length is a = 1, the focus is (2,1), and the y-axis is the directrix. y 1 S(2,1) 0 1 x Exercise 14 The line 3x + 2y = 1. Exercise 15 If PQ is a focal chord, then pq = −1. Hence we can write the coordinates of Q as ( − 2a p , a p 2 ) . Thus ( PQ 2 = 2ap + 2a ) 2 + (ap 2 − a ) 2 p p 2 = 4a 2 p 2 + 8a 2 + 4a2 p 2 = a 2( p 4 + 4p 2 + 6 + 4 p 2 + 1 p 4 ) = a 2( p + 1 p ) 4, + a2 p 4 − 2a 2 + a2 p 4 where we recall the numbers 1,4,6,4,1 from Pascal’s triangle and the binomial theorem. ( So PQ = a p + 1 ) 2. p
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Quadratics - the Australian Mathematical Sciences Institute

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