Question

Pls help with my math plssss

Answer 1

Consider the figure posted below.

The cars start from the common point A. The car heading east travels a certain distance, say
`x`, and reaches point B. The car heading south travels 10 miles more, so
`x+10`, and reaches point C. The distance between the two cars, BC, is 50.

Since ABC is a right triangle, we can apply Pythagorean's theorem:

`AB^2+AC^2=BC^2`

Plug the expressions/value for each side:

`x^2+(x+10)^2=50^2`

Expand/compute the squares:

`x^2+x^2+20x+100 = 2500 \iff 2x^2+20x-2400 = 0 \iff x^2+10x-1200=0`

The two solutions of this equation are
`x=40,\ x=-30`

Since the cars can't travel a negative distance, the only acceptable solution is
`x=40`