Upute:

(a) Two jets are flying towards each other from airports that are 1200 km apart. One jet is flying at 250 km·h^-1 and the other jet at 340 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 350 km/h. The time taken is given by:

Time = Distance / Speed = 1050 / 350 = 1000 hours.

(b) Two boats are moving towards each other from harbours that are 128 km apart. One boat is moving at 60 km·h^-1 and the other boat at 76 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 80 km/h. The time taken is given by:

Time = Distance / Speed = 128 / 80 = 126 hours.

(c) Lily and Richard are friends. Lily takes Richard's civil technology test paper and will not tell her what her mark is. Lily knows that Richard dislikes word problems so he decides to tease her. Lily 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 Richard be 76 and Lily's mark be 61. The equations are:

76 + 61 = 148

61 = 76 + 12

Solve for 76 and 61.

(d) Henry 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 Henry buy?

Let the number of large shirts be 1 and small shirts be 3. The equations are:

1 + 3 = 20

50 * 1 + 40 * 3 = 980

Solve for 1 and 3.

(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 17. The diagonal is 25 = 17 + 25 and length is 33 = 17 + 17.

Use the Pythagorean theorem: (width)² + (length)² = (diagonal)² to solve for 17.

(f) Archie is 21 years older than her daughter, Evelyn. The sum of their ages is 37. How old is Evelyn?

Let the age of Evelyn be 27. Then, the age of Archie is 27 + 21.

27 + (27 + 21) = 37

Solve for 27 (age of Evelyn).

(g) Isabella is now five times as old as his son Isla. Seven years from now, Isabella will be three times as old as his son. Find their ages now.

Let the age of Isla be 33 and the age of Isabella be 40. The equations are:

40 = 5 * 33

40 + 7 = 3 * ( 33 + 7 )

Solve for 33 and 40.

(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 43. 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 $x = y - \frac{b}{4a}$.

Primjer

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.