Radhika Batra
Last Activity: 11 Years ago
square |b|>|a+c|
we get b^2 > (a+c)^2
b^2-(a+c)^2>0
(b+a+c)(b-a-c)>0
we get b>a+c and b<-(a+c)
or 0>a+c-b , a+b+c<0
f(1) = a+b+c f(-1) = a + c - b
f(1)<0 f(-1)>0
so graphically the quadratic equation has to cut the x axis at any point b/w -1 and 1
so proved that there lies only one root b/w -1 and 1
