Given that in a quadratic equation $ax^2+ bx + c=0$, $(4a+c)^2<4b^2$, Find the interval in which roots lie.
I subtracted $16ac$ from both sides, to get $\Delta>(4a-c)^2$, which is always greater than zero, hence roots are real. but otherwise, I haven't been able to do anything substantial. Any help will be appreciated.
Thanks in advance!!
Using $(4a+c)^2<(2b)^2\Rightarrow -2b< 4a+c<2b$
So we get $4a+2b+c>0$ and $4a-2b+c<0$
Now Here $f(x)=ax^2+bx+c\;,$ Put $x=-2\;,$ We get $f(-2)=4a-2b+c<0$
and Put $x=+2\;,$ We get $f(2) = 4a+2b+c>0$
So $f(-2)<0$ and $f(2)>0$
Now Yoc can Solve it.