Question

From the same place at 7am 'A' started walking in north at the speed of 4 km/hr. After 1 hour B started cycling in the east at the speed of 8 km/hr. At what time will they be at a distance of 20 km apart from each other????

Answer 1

we have,
$80 {x }^{2} - 128x - 336 = 0$

$(dividing \: throughout \: by \: 16)$

$5 {x}^{2} - 8x - 21 = 0$
by solving quadratic equation by factorization method.

$5 {x}^{2} - 15x + 7x - 21 = 0$

THEREFORE,
$5x(x - 3) + 7(x - 3) = 0$

$(x - 3)(5x + 7) = 0$


$x - 3 = 0 \: or \: 5x + 7 = 0$

$x = 3 \: or \: x = \frac{ - 7}{5}$

$but \: x = \frac{ - 7}{5} \:is \: not \: acceptable$

$therefore \: x = 3$

THUS,AFTER 3 hours from 7 a.m A & B are 20 km apart from each other.

$Final \: answer \\ \: i.e \: at \: 10a.m \: they \: are 20km \: \\ apart \: from \: each \: other$


$hope \: this \: helps \: you$

Answer 2

we have,

80 {x }^{2} - 128x - 336 = 0

(dividing \: throughout \: by \: 16)

5 {x}^{2} - 8x - 21 = 0

by solving quadratic equation by factorization method.

5 {x}^{2} - 15x + 7x - 21 = 0

THEREFORE,

5x(x - 3) + 7(x - 3) = 0

(x - 3)(5x + 7) = 0

x - 3 = 0 \: or \: 5x + 7 = 0

x = 3 \: or \: x = \frac{ - 7}{5}

but \: x = \frac{ - 7}{5} \:is \: not \: acceptable

therefore \: x = 3

THUS,AFTER 3 hours from 7 a.m A & B are 20 km apart from each other.

Final \: answer \\ \: i.e \: at \: 10a.m \: they \: are 20km \: \\ apart \: from \: each \: other

Read more on Brainly.in - https://brainly.in/question/2581538#readmore

Answer:

Step-by-step explanation: