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<details> <summary>Upute: </summary> <img src="https://www.mathkiss.com/uploads/hypo5.webp" width="500"/> </details> (a) Two jets are flying towards each other from airports that are §§V1(1000,1200,50)§§ km apart. One jet is flying at §§V2(240,250,10)§§ km·h-1 and the other jet at §§V3(340,350,10)§§ km·h-1. If they took off at the same time, how long will it take for the jets to pass each other? The relative speed is §§V4(240,350,10)§§ km/h. The time taken is given by: Time = Distance / Speed = §§V5(1000,1200,50)§§ / §§V4(240,350,10)§§ = §§V6(1000,1200,50)§§ hours. (b) Two boats are moving towards each other from harbours that are §§V7(120,144,2)§§ km apart. One boat is moving at §§V8(60,65,1)§§ km·h-1 and the other boat at §§V9(75,80,1)§§ km·h-1. If both boats started their journey at the same time, how long will they take to pass each other? The relative speed is §§V10(60,80,1)§§ km/h. The time taken is given by: Time = Distance / Speed = §§V7(120,144,2)§§ / §§V10(60,80,1)§§ = §§V11(120,144,2)§§ hours. (c) §§N1§§ and §§N2§§ are friends. §§N1§§ takes §§N2§§'s civil technology test paper and will not tell her what her mark is. §§N1§§ knows that §§N2§§ dislikes word problems so he decides to tease her. §§N1§§ says: “I have 12 marks more than you do and the sum of both our marks is equal to 148. What are our marks?” Let the mark of §§N2§§ be §§V12(50,148,1)§§ and §§N1§§'s mark be §§V13(50,148,1)§§. The equations are: §§V12(50,148,1)§§ + §§V13(50,148,1)§§ = 148 §§V13(50,148,1)§§ = §§V12(50,148,1)§§ + 12 Solve for §§V12(50,148,1)§§ and §§V13(50,148,1)§§. (d) §§N3§§ bought 20 shirts at a total cost of R 980. If the large shirts cost R 50 and the small shirts cost R 40, how many of each size did §§N3§§ buy? Let the number of large shirts be §§V14(0,20,1)§§ and small shirts be §§V15(0,20,1)§§. The equations are: §§V14(0,20,1)§§ + §§V15(0,20,1)§§ = 20 50 * §§V14(0,20,1)§§ + 40 * §§V15(0,20,1)§§ = 980 Solve for §§V14(0,20,1)§§ and §§V15(0,20,1)§§. (e) The diagonal of a rectangle is 25 cm more than its width. The length of the rectangle is 17 cm more than its width. What are the dimensions of the rectangle? Let the width be §§V16(1,20,1)§§. The diagonal is §§V17(25,50,25)§§ = §§V16(1,20,1)§§ + 25 and length is §§V18(17,35,1)§§ = §§V16(1,20,1)§§ + 17. Use the Pythagorean theorem: (width)2 + (length)2 = (diagonal)2 to solve for §§V16(1,20,1)§§. (f) §§N4§§ is 21 years older than her daughter, §§N5§§. The sum of their ages is 37. How old is §§N5§§? Let the age of §§N5§§ be §§V19(10,37,1)§§. Then, the age of §§N4§§ is §§V20(10,37,1)§§ + 21. §§V19(10,37,1)§§ + (§§V19(10,37,1)§§ + 21) = 37 Solve for §§V19(10,37,1)§§ (age of §§N5§§). (g) §§N6§§ is now five times as old as his son §§N7§§. Seven years from now, §§N6§§ will be three times as old as his son. Find their ages now. Let the age of §§N7§§ be §§V21(1,50,1)§§ and the age of §§N6§§ be §§V22(5,50,5)§§. The equations are: §§V22(5,50,5)§§ = 5 * §§V21(1,50,1)§§ §§V22(5,50,5)§§ + 7 = 3 * ( §§V21(1,50,1)§§ + 7 ) Solve for §§V21(1,50,1)§§ and §§V22(5,50,5)§§. (h) If adding one to three times a number is the same as the number, what is the number equal to? Let the number be §§V23(1,50,1)§§. The equation is: Opći oblik kvartične jednadžbe je: $$ax^4 + bx^3 + cx^2 + dx + e = 0$$ Supstitucijom $x = y - \frac{b}{4a}$ dobivamo reducirani oblik: $$y^4 + py^2 + qy + r = 0$$ gdje su $p$, $q$ i $r$ izraženi preko $a$, $b$, $c$, $d$ i $e$. Ferrarijeva rezolventa je: $$z^3 + 2pz^2 + (p^2 - 4r)z - q^2 = 0$$ Nakon rješavanja kubične rezolvente (npr. Cardanovom metodom) i dobivanja rješenja $z_0$, rješenja reducirane jednadžbe su: $$y = \pm \sqrt{-\frac{p}{2} \pm \sqrt{\left(\frac{p}{2}\right)^2 - r + \frac{z_0}{2}}}$$ Konačno, vraćamo se na $x$ koristeći $x = y - \frac{b}{4a}$. <h2>Primjer</h2> Riješimo jednadžbu: $$x^4 - 10x^3 + 35x^2 - 50x + 24 = 0$$ Supstitucijom $$x = y + \frac{10}{4} = y + 2.5$$ dobivamo: $$y^4 - 5y^2 + 4 = 0$$ Dakle, $$p = -5$$, $$q = 0$$, $$r = 4$$. Ferrarijeva rezolventa je: $$z^3 - 10z^2 + 9z = 0$$ što se može faktorizirati u $z(z-1)(z-9) = 0$. Uzmimo $z_0 = 1$. Rješenja za $y$ su: $$y = \pm \sqrt{\frac{5}{2} \pm \sqrt{\frac{25}{4} - 4 + \frac{1}{2}}} = \pm \sqrt{2.5 \pm \sqrt{2.75}}$$ $$y \approx \pm 2.04, \pm 0.92$$ Konačno, rješenja za $x$ su: $$x = y + 2.5$$ $$x \approx 4.54, 0.46, 3.42, 1.58$$ Točna rješenja su $$x=1, 2, 3, 4$$, što se može dobiti faktorizacijom $$(x-1)(x-2)(x-3)(x-4) = 0$$. Razlika je zbog zaokruživanja u računanju korijena.