root a, root b, root c are sides of a triangle. Find no. of solutions(real) in x, ax^2+(a+b-c)x+b=0

Asked by aahnik.mohanty | 21st Jun, 2019, 09:00: PM

Expert Answer:

Since space square root of straight a comma square root of straight b space and space square root of straight c space are space sides space of space triangle space they space all space are space greater than 0
straight a comma straight b comma straight c greater than 0
ax squared plus open parentheses straight a plus straight b minus straight c close parentheses straight x plus straight b equals 0
roots space are space given space by
fraction numerator negative open parentheses straight a plus straight b minus straight c close parentheses plus-or-minus square root of open parentheses straight a plus straight b minus straight c close parentheses squared minus 4 ab end root over denominator 2 straight a end fraction
for space real space roots space
open parentheses straight a plus straight b minus straight c close parentheses squared minus 4 ab space should space be space greater than 0
rightwards double arrow straight a squared plus straight b squared plus straight c squared plus 2 ab minus 2 bc minus 2 ac minus 4 ab
rightwards double arrow straight a squared plus straight b squared plus straight c squared minus 2 ab minus 2 bc minus 2 ac

Check space your space question space again space more space info space is space needed space to space proceed space further

Answered by Arun | 29th Jun, 2019, 09:36: AM