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    While going through a physics problem, I found following mathematical property being used.

    If a > 0 and b2 - 4ac < 0, then ax2 + bx + c > 0 for all x.

    Can anyone explain, why the property holds true?
    Thanks in advance.
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    (Original post by tangotangopapa2)
    While going through a physics problem, I found following mathematical property being used.

    If a > 0 and b2 - 4ac < 0, then ax2 + bx + c > 0 for all x.

    Can anyone explain, why the property holds true?
    Thanks in advance.
    Are you familiar with the formula for the roots of a quadratic equation?
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    (Original post by ghostwalker)
    Are you familiar with the formula for the roots of a quadratic equation?
    Yes. I am.
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     (\Delta =) b^2-4ac is called the disriminant of the quadratic equation, it tells us of the nature of the roots, if  \Delta &lt;0 then the equation has complex roots (non real), if  \Delta=0 then the equation has a repeated root (one root basically where the curve is tangent to the x-axis) and if  \Delta &gt;0 then the equation has 2 distinct real roots.
    The general solution to the quadratic you gave is  \displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a} . 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 x-axis or not.
    If  \Delta &lt; 0 then the curve y does not cross the x-axis, 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  b^2-4ac =-9 , what would the value of x be? Certainly no real number, so it cannot cross the x axis right.
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    (Original post by tangotangopapa2)
    Yes. I am.
    Good, that saves a lot of LaTex.

    b2 - 4ac < 0 implies tha the quadratic has no real roots, i.e. it does not cross the x-axis. So, it is always positive or always negative.

    If, in addition, a>0, then as x goes off to infinity, so ax2 + 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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    Well, first, that's incorrect - if b2 - 4ac < 0 then there's no real solution of x.

    b2 - 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 ax2 + 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

    (b2/4a2) - c/a 0

    (b2/4a2) - 4ac/4a2 0

    (b2 - 4ac)/4a2 0

    And therefore, as 4a2 is always positive, b2 - 4ac 0 for there to be real solutions of x.
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    (Original post by B_9710)
     (\Delta =) b^2-4ac is called the disriminant of the quadratic equation, it tells us of the nature of the roots, if  \Delta &lt;0 then the equation has complex roots (non real), if  \Delta=0 then the equation has a repeated root (one root basically where the curve is tangent to the x-axis) and if  \Delta &gt;0 then the equation has 2 distinct real roots.
    The general solution to the quadratic you gave is  \displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a} . 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 x-axis or not.
    If  \Delta &lt; 0 then the curve y does not cross the x-axis, 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  b^2-4ac =-9 , what would the value of x be? Certainly no real number, so it cannot cross the x axis right.
    Thank you so much. Got it.
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    (Original post by ghostwalker)
    Good, that saves a lot of LaTex.

    b2 - 4ac < 0 implies tha the quadratic has no real roots, i.e. it does not cross the x-axis. So, it is always positive or always negative.

    If, in addition, a>0, then as x goes off to infinity, so ax2 + 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.
    Thanks alot.
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    (Original post by Alexion)
    Well, first, that's incorrect - if b2 - 4ac < 0 then there's no real solution of x.

    b2 - 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 ax2 + 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

    (b2/4a2) - c/a 0

    (b2/4a2) - 4ac/4a2 0

    (b2 - 4ac)/4a2 0

    And therefore, as 4a2 is always positive, b2 - 4ac 0 for there to be real solutions of x.
    Thank you very much.
    But question does not ask for equality. It asks for inequality.
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    (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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    There is also the following formula  \displaystyle \frac{-2c}{b\pm\sqrt{b^2-4ac}} that no-one uses.
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    (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?
    Yes, one implies the other.
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    (Original post by Student403)
    ???

    It's simple - there's just one. Not 2 identical ones
    All quadratic equations have two solutions as a fundamental rule - so both solutions are the same :yep:
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    (Original post by Ano123)
    There is also the following formula  \displaystyle \frac{-2c}{b\pm\sqrt{b^2-4ac}} that no-one uses.
    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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    (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
    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.
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    (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.
    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 n-roots, no more.

    Anyway thanks.
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    (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 n-roots, no more.

    Anyway thanks.
    You would have to take the positive root of the b^2 because otherwise you would get 0 on the denominator. It gives x=-c/b which of course is the solution to bx+c=0.
 
 
 
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