Question

PRACTISE NOW 11

Answer 1

Answer:

i. x² + 7x - 16800 = 0 ii. x = 126.16 km/h or -133.16 km/h iii. 5.01 h

Step-by-step explanation:

i. If he drove at an average speed of x km/h on his journey from City P to City Q formulate an equation in x and show that it reduces to x2 + 7x – 16 800 = 0.

For the first journey from City P to City Q, with Mr Lee moving at an average speed of x km/h, he reaches there in time, t and covers the distance, d = 600 km

So, xt = 600 (1)

On his return journey from City Q to CIty P, his average speed increases by 7 km/h, so it is (x + 7)km/h and his time is 15 minutes less than his first journey. 15 min = 15/60 h = 0.25 h, we have that his time for the journey is (t - 0.25) h. Since the distance covered is the same d = 600 km,

We have (x + 7)(t - 0.25) = 600  (2)

Expanding the brackets, we have

xt - 0.25x + 7t - 0.25(7) = 600

xt - 0.25x + 7t - 1.75 = 600

From (1) t = 600/x and xt = 600

Substituting these into the equation, we have

600 - 0.25x + 7(600/x) - 1.75 = 600

simplifying

-0.25x + 4200/x - 1.75 = 600 - 600

-0.25x + 4200/x - 1.75 = 0

multiplying through by x, we have

-0.25x² + 4200 - 1.75x = 0

dividing through by -0.25, we have

-0.25x²/-0.25 + 4200/-0.25 - 1.75x/-0.25 = 0

x² - 16800 + 7x = 0

re-arranging, we have

x² + 7x - 16800 = 0

ii. Solve the equation x² + 7x - 16 800 = 0, giving both your answers correct to  2 decimal places.

Using the quadratic formula, we solve x² + 7x - 16800 = 0 for x

So, $x = \frac{-7 +/-\sqrt{7^{2} - 4 X 1 X -16800} }{2 X 1}\\x = \frac{-7 +/-\sqrt{49 + 67200} }{2} \\x = \frac{-7 +/-\sqrt{67249} }{2} \\x = \frac{-7 +/- 259.32}{2} \\x = \frac{-7 + 259.32}{2} or x = \frac{-7 - 259.32}{2} \\x = 252.32/2 or x= -266.32/2\\x = 126.16 km/hor x = -133.16 km/h$

So, x = 126.16 km/h or -133.16 km/h

iii. Find the time taken for the return journey

The time taken for the return journey is t' = t + 0.25. Now. t = 600/x

Since x cannot be negative, we use x = 126.16 km/h.

So, t = 600/x = 600/126.16 = 4.76 h

t' = t + 0.25

t' = 4.76 + 0.25

t' = 5.01 h