While going through a physics problem, I found following mathematical property being used.
If a > 0 and b^{2}  4ac < 0, then ax^{2} + bx + c > 0 for all x.
Can anyone explain, why the property holds true?
Thanks in advance.
Quadratic Equation Help
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 05072016 13:34
Last edited by tangotangopapa2; 05072016 at 13:35. 
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 05072016 13:41
(Original post by tangotangopapa2)
While going through a physics problem, I found following mathematical property being used.
If a > 0 and b^{2}  4ac < 0, then ax^{2} + bx + c > 0 for all x.
Can anyone explain, why the property holds true?
Thanks in advance. 
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 3
 05072016 13:43
(Original post by ghostwalker)
Are you familiar with the formula for the roots of a quadratic equation? 
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 05072016 13:46
is called the disriminant of the quadratic equation, it tells us of the nature of the roots, if then the equation has complex roots (non real), if then the equation has a repeated root (one root basically where the curve is tangent to the xaxis) and if then the equation has 2 distinct real roots.
The general solution to the quadratic you gave is . So if the expression inside the square root is negative then it cannot possibly be a real number. Basically the discriminant tells us the curve crosses the xaxis or not.
If then the curve y does not cross the xaxis, if a>0 this means that the curve lies completely above the x axis. If a<0 it means that the curve is always under the x axis.
For example if is , what would the value of x be? Certainly no real number, so it cannot cross the x axis right.Last edited by B_9710; 05072016 at 13:49.Post rating:3 
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 05072016 13:50
(Original post by tangotangopapa2)
Yes. I am.
b^{2}  4ac < 0 implies tha the quadratic has no real roots, i.e. it does not cross the xaxis. So, it is always positive or always negative.
If, in addition, a>0, then as x goes off to infinity, so ax^{2} + bx + c goes off to infinity, i.e. it's positive from some point onwards.
Putting the two together, since it must be positive at some point, it is always positive.Last edited by ghostwalker; 05072016 at 13:52.Post rating:1 
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 05072016 13:51
Well, first, that's incorrect  if b^{2}  4ac < 0 then there's no real solution of x.
b^{2}  4ac is known as the discriminant of the quadratic. If < 0, there's real solutions, if = 0, there's two identical solutions (one, for all intents and purposes), and if > 0, then there's two different real solutions of the quadratic.
It's shown as part of the quadratic formula
where it's evident that it cannot be solved if the discriminant is less than zero.
It's actually derived by completing the square. For the equation ax^{2} + bx + c = 0:
x^2 + (b/a)x + c/a = 0
(x + b/2a)^{2}  (b/2a)^{2} + c/a = 0
(x + b/2a)^{2 }= (b/2a)^{2}  c/a
As the LHS is a square:
(b/2a)^{2}  c/a ≥ 0
(b^{2}/4a^{2})  c/a ≥ 0
(b^{2}/4a^{2})  4ac/4a^{2}≥ 0
(b^{2}  4ac)/4a^{2}≥ 0
And therefore, as 4a^{2} is always positive, b^{2}  4ac ≥ 0 for there to be real solutions of x.Post rating:1 
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 05072016 13:53
(Original post by B_9710)
is called the disriminant of the quadratic equation, it tells us of the nature of the roots, if then the equation has complex roots (non real), if then the equation has a repeated root (one root basically where the curve is tangent to the xaxis) and if then the equation has 2 distinct real roots.
The general solution to the quadratic you gave is . So if the expression inside the square root is negative then it cannot possibly be a real number. Basically the discriminant tells us the curve crosses the xaxis or not.
If then the curve y does not cross the xaxis, if a>0 this means that the curve lies completely above the x axis. If a<0 it means that the curve is always under the x axis.
For example if is , what would the value of x be? Certainly no real number, so it cannot cross the x axis right. 
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 05072016 13:54
(Original post by ghostwalker)
Good, that saves a lot of LaTex.
b^{2}  4ac < 0 implies tha the quadratic has no real roots, i.e. it does not cross the xaxis. So, it is always positive or always negative.
If, in addition, a>0, then as x goes off to infinity, so ax^{2} + bx + c goes off to infinity, i.e. it's positive from some point onwards.
Putting the two together, since it must be positive at some point, it is always positive. 
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 05072016 13:56
(Original post by Alexion)
Well, first, that's incorrect  if b^{2}  4ac < 0 then there's no real solution of x.
b^{2}  4ac is known as the discriminant of the quadratic. If < 0, there's real solutions, if = 0, there's two identical solutions (one, for all intents and purposes), and if > 0, then there's two different real solutions of the quadratic.
It's shown as part of the quadratic formula
where it's evident that it cannot be solved if the discriminant is less than zero.
It's actually derived by completing the square. For the equation ax^{2} + bx + c = 0:
x^2 + (b/a)x + c/a = 0
(x + b/2a)^{2}  (b/2a)^{2} + c/a = 0
(x + b/2a)^{2 }= (b/2a)^{2}  c/a
As the LHS is a square:
(b/2a)^{2}  c/a ≥ 0
(b^{2}/4a^{2})  c/a ≥ 0
(b^{2}/4a^{2})  4ac/4a^{2}≥ 0
(b^{2}  4ac)/4a^{2}≥ 0
And therefore, as 4a^{2} is always positive, b^{2}  4ac ≥ 0 for there to be real solutions of x.
But question does not ask for equality. It asks for inequality.Last edited by tangotangopapa2; 05072016 at 13:59. 
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 05072016 13:57
(Original post by Alexion)
if = 0, there's two identical solutions (one, for all intents and purposes)
???
It's simple  there's just one. Not 2 identical ones 
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 05072016 13:58
Post rating:1 
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 05072016 13:59
(Original post by tangotangopapa2)
Thank you very much.
Is the converse also true?
i.e if the equation is positive for all values of x then, a>0 and discriminant<0? 
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 05072016 14:00
Post rating:1 
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 05072016 14:05
I'm gonna use it where ever possible . :d 
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 05072016 14:09
(Original post by tangotangopapa2)
Cool. Took the ratio of two formulas and ans is 1. Verified.
I'm gonna use it where ever possible . :d 
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 05072016 14:16
(Original post by Ano123)
They are just really the same formula, just rationalise the denominator to get to the more conventional one. Advantage of this one though is that it gives you a solution even if a=0.
Anyway thanks. 
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 05072016 14:19
(Original post by tangotangopapa2)
Ah. It falsely gives you two roots for linear equation. Though the second root is undefined, does it not violate one of the fundamental theorems of algebra? A nth degree equation has nroots, no more.
Anyway thanks.
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Updated: July 5, 2016
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