1. The minimum number of roads joining 4 towns to each other is 6 as shown. State the minimum number of roads needed to join 7 towns to each other.

Transcription

1 Homework 6 (Applying the content Quadratics) 1. The minimum number of roads joining 4 towns to each other is 6 as shown. The minimum number of roads, r, joining n towns to each other is given by the formula r = 1 n(n 1). 2 State the minimum number of roads needed to join 7 towns to each other. (b) When r = 55, show that n 2 n 110 = 0. (c) Hence find algebraically the value of n. r = 1 7 (7 1) sub for n the number of towns 2 Min 21 roads process the correct answer 55 = 1 n(n 1) sub for 55 roads = n 2 n eliminate the fraction n 2 n 110 = 0 as required process to answer (n 11)(n + 10) = 0 factorise LHS n = 11 or n = 10 for both solutions to the quadratic equation n = 10 rejecting the negative answer.

2 2. A rectangular garden has a length of (x + 7) metres and a breadth of (x + 3) metres. (b) Show that the area, A square metres, of the garden is given by A = x x If the area of the garden is 45 square metres, find x. A = (x + 7)(x + 3) sub into area of rectangle formula A = x x + 21 process the correct answer 45 = x x + 21 sub for area of 45 square metres x x 24 = 0 set the quadratic equal to zero (x + 12)(x 2) = 0 factorising the quadratic equation x = 2 or x = 12 for both solutions to the quadratic equation x = 12 rejecting the negative answer. 3. A right angled triangle has dimensions, in centimetres, as shown. x x + 8 Calculate the value of x. x + 7 (x + 8) 2 = x 2 + (x + 7) 2 apply Pythagoras x x + 64 = x 2 + x x + 49 expand both brackets x 2 2x 15 = 0 set the quadratic equal to zero (x + 3)(x 5) = 0 factorising the quadratic equation x = 3 or x = 5 for both solutions x = 3 rejecting the negative answer.

3 4. The weight, W kilograms, of a giraffe is related to its age, M months, by the formula W = 1 4 (M2 4M + 272). At what age will the giraffe weigh 83 kilograms? 83 = 1 4 (M2 4M + 272) substitute the weight 332 = M 2 4M eliminate the fraction M 2 4M 60 = 0 set the quadratic equal to zero (M + 6)(M 10) = 0 factorising the quadratic equation M = 6 or M = 10 for both solutions M = 6 rejecting the negative answer. 5. The profit made by a publishing company of a magazine is calculated by the formula y = 4x(140 x) where y is the profit (in pounds) and x is selling price (in pence) of the magazine. The graph below represents the profit y against the selling price x. y 0 Find the maximum profit the company can make from the sale of the magazine. x 0 = 4x(140 x) cuts x-axis when y = 0. x = 0, x = 140 finds the roots. max profit at x = 70 axis of symmetry at midpoint of roots y = 4 70 (140 70) = max profit on the axis of symmetry

4 6. The diagram below shows the path of a rocket which is fired into the air. The height, h metres, of the rocket after t seconds is given by h(t) = 2t(t 14). h For how many seconds is the rocket in flight? (b) What is the maximum height reached by the rocket? 0 t 0 = 2t(t 14) cuts x-axis when y = 0 t = 0, t = 14 finds the roots Rocket in the air for 14 seconds communicate solution max height at t = 7 axis of symmetry at midpoint of roots h = 2 (7 14) = 98m max height on the axis of symmetry 7. A decorator s logo is rectangular and measures 10 centimetres by 6 centimetres. It consists of three rectangles: one red, one yellow and one blue. The yellow rectangle measures 10 centimetres by x centimetres. The width of the red rectangle is x centimetres. Show that the area, A, of the blue rectangle is given by the expression A = x 2 16x (b) The area of the blue rectangle is equal to 1 of the total area of the logo. 5 Calculate the value of x.

5 A = (10 x)(6 x) sub into area of rectangle formula A = 60 16x + x 2 A = x 2 16x + 60 expanding the brackets and completing proof A logo = 1 60 = 5 12m2 Area of logo x 2 16x + 60 = 12 Sub area of logo into formula from x 2 16x + 48 = 0 set the quadratic equal to zero (x 4)(x 12) = 0 factorising the quadratic equation x = 4 or x = 12 for both solutions to the quadratic equation x = 12 for rejecting 12 as too large. 8. Given that ax 2 + 4x 2 = 0 has equal roots. Find a. b 2 4ac = 0 for equal roots strategy of using the discriminant 16 4 a ( 2) = a = 0 apply the discriminant a = 2 process to the solution 9. It is known that the equation x 2 + bx + 25 = 0 has 1 root. Find 2 values for b. b 2 4ac = 0 for equal roots strategy of using the discriminant b = 0 apply the discriminant b = ±10 For both solutions

6 10. The equation px 2 + 8x 2 = 0 has 2 real roots. Set up an inequality in p, and solve it. b 2 4ac > 0 for 2 real roots strategy of using the discriminant 64 4 p ( 2) > p > 0 8p > 64 apply the discriminant p > 8 process to the solution 11. Given that x 2 + x t = 0 has no real roots. Find the value of t. b 2 4ac < 0 for no real roots strategy of using the discriminant ( t) < t < 0 4t < 1 apply the discriminant p < 1 4 process to the solution