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If $a,b,c$ are real numbers and $a+b+c =0$ then how to prove that the equation $$4ax^2+3bx+2c$$ has two real roots. I just know that for real roots the quadratic equation should have its Discriminant greater than or equal to zero but how can is use the condition that $a+b+c=0$ . Any hint might help .Thanks

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Guide:

The discriminant is

\begin{align}(3b)^2-4(4a)(2c)&=9b^2-32ac \\ &=9b^2+32(b+c)c\end{align}

You might like to do a completing the square to show that it is positive.

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Alternatively, the discriminant is $$9b^2-32a(-a-b)=b^2+8(2a+b)^2>0,$$ since $a\neq 0$ and at least one of $b\neq 0,\; b\neq -2a$ holds.

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