Upute:

(a) Two jets are flying towards each other from airports that are 1000 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 300 km/h. The time taken is given by:

Time = Distance / Speed = 1100 / 300 = 1100 hours.

(b) Two boats are moving towards each other from harbours that are 138 km apart. One boat is moving at 63 km·h^-1 and the other boat at 77 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 63 km/h. The time taken is given by:

Time = Distance / Speed = 138 / 63 = 134 hours.

(c) John and Elijah are friends. John takes Elijah's civil technology test paper and will not tell her what her mark is. John knows that Elijah dislikes word problems so he decides to tease her. John 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 Elijah be 77 and John's mark be 126. The equations are:

77 + 126 = 148

126 = 77 + 12

Solve for 77 and 126.

(d) Noah 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 Noah buy?

Let the number of large shirts be 17 and small shirts be 20. The equations are:

17 + 20 = 20

50 * 17 + 40 * 20 = 980

Solve for 17 and 20.

(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 20. The diagonal is 50 = 20 + 25 and length is 24 = 20 + 17.

Use the Pythagorean theorem: (width)² + (length)² = (diagonal)² to solve for 20.

(f) Grace 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 15. Then, the age of Grace is 17 + 21.

15 + (15 + 21) = 37

Solve for 15 (age of Evelyn).

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

Let the age of Oscar be 24 and the age of Liam be 10. The equations are:

10 = 5 * 24

10 + 7 = 3 * ( 24 + 7 )

Solve for 24 and 10.

(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 11. 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.