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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!!

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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.

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  • $\begingroup$ Thanks a lot @juantheron .... I only needed the interval, so I don't really have to solve it now...the interval is (-2,2)...thanks $\endgroup$ – GRrocks Oct 4 '15 at 11:29

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