I was solving a problem in physics when I encountered this problem. I wanted to find the condition where the value of a quadratic equation is positive. Suppose the function is : $f(x) = Ax^2+Bx+C$
Then, I need to find the condition for which the function will give me positive value.Please Help!
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Bernard
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therealyubraj
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1There are two conditions : $A >0$ and $B^2-4AC<0$. They must be both satisfied. – Crostul Feb 15 '18 at 13:21
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Yeah I came across this but what i need to know is how to derive these conditions. – therealyubraj Feb 15 '18 at 13:22
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See https://www.khanacademy.org/math/algebra-home/alg-quadratics/alg-quadratic-inequalities/v/quadratic-inequalities-visual-explanation – Crostul Feb 15 '18 at 13:28
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We can express
$$y = ax^2 + bx +c= \frac{(2ax+b)^2 +4ac - b^2}{4a}$$
Now it is clear, $y$ is positive when $4a$ is positive and $4ac - b^2$ is positive.
Especially Lime
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nimmy
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2Hey, I formatted your answer in MathJax for you (the edit will be visible shortly) I suggest you read a basic MathJax guide as it is the standard for math formatting here and it makes math much more readable! Have a nice day – user438666 Feb 15 '18 at 13:38
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Let $x_1,x_2$ be the solutions of $f(x)=0$.
Case 1: $x_1,x_2 \notin \mathbb R$. If $A>0$, then $f(x) >0$ for all $x \in \mathbb R.$ If $A<0$, then $f(x) <0$ for all $x \in \mathbb R.$
Case 2: $x_1,x_2 \in \mathbb R$. We can assume that $x_1 \le x_2$. If $A>0$ , then $f(x)>0$ $ \iff x<x_1$ or $x>x_2$. It is now your turn to investigate the situation if $A<0$.
Fred
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